How to show that $gcd(a_1,a_2,cdots,a_k) = 1$ implies that there exist a non-negative solution to...












4












$begingroup$


I was reading about the Coin-problem and I am unable to fully understand the following argument:




On the other hand, whenever the GCD equals 1, the set of integers that cannot be expressed as a conical combination of ${ a_1, a_2, …, a_n }$ is bounded according to Schur's theorem, and therefore the Frobenius number exists.




Here the author is arguing for the existence of a non-negative solution to the linear Diophantine equation (LDE)
$$sum_{i=1}^{k}a_ix_i =n text{}$$
for large enough $n.$ Now I tried to understand the proof of the Schur theorem here enter link description here (Page 98), but I am not sure that I understand it fully. In particular, I don't understand why the generating function associated with the sequence $h_n$ that counts the number of solutions to the LDE is
$$H(x)= prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right).$$



Once we have $H(x)$ we can deduce that
$$h_nsim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$ How does this exactly show that for large enough $n,$ $h_n>0?$ Is it because the $frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}>0?$










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  • 2




    $begingroup$
    Yes.$phantom{}$
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:49






  • 3




    $begingroup$
    Consider just the case $k=2,a_1=3,a_2=7$ for clarity. Can you see that the coefficient of $x^n$ in $$(1+x^3+x^6+x^9+ldots)(1+x^7+x^{14}+x^{21}+ldots)$$ is exactly the number of ways for writing $n$ as $3A+7B$?
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:51








  • 1




    $begingroup$
    Well, by geometric series $(1+x^3+x^6+ldots)=frac{1}{1-x^3}$ and $(1+x^7+x^{14}+ldots)=frac{1}{1-x^7}$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:53






  • 2




    $begingroup$
    The meromorphic function $frac{1}{(1-x^3)(1-x^7)}$ has a double pole at $x=1$ but the other singularities are all simple poles. Any simple pole provides a bounded contribution, since the coefficients of $frac{1}{1-xxi}$ have unit modulus for $xiin S^1$, hence the magnitude of the representation function is linear, since $[x^n]frac{1}{(1-x)^2}=Theta(n)$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:55








  • 1




    $begingroup$
    Linear and with a bounded perturbation implies strictly positive from some point on - the hidden constant in $Theta$ is not really important, but it can be derived from stars and bars.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:56


















4












$begingroup$


I was reading about the Coin-problem and I am unable to fully understand the following argument:




On the other hand, whenever the GCD equals 1, the set of integers that cannot be expressed as a conical combination of ${ a_1, a_2, …, a_n }$ is bounded according to Schur's theorem, and therefore the Frobenius number exists.




Here the author is arguing for the existence of a non-negative solution to the linear Diophantine equation (LDE)
$$sum_{i=1}^{k}a_ix_i =n text{}$$
for large enough $n.$ Now I tried to understand the proof of the Schur theorem here enter link description here (Page 98), but I am not sure that I understand it fully. In particular, I don't understand why the generating function associated with the sequence $h_n$ that counts the number of solutions to the LDE is
$$H(x)= prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right).$$



Once we have $H(x)$ we can deduce that
$$h_nsim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$ How does this exactly show that for large enough $n,$ $h_n>0?$ Is it because the $frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}>0?$










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes.$phantom{}$
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:49






  • 3




    $begingroup$
    Consider just the case $k=2,a_1=3,a_2=7$ for clarity. Can you see that the coefficient of $x^n$ in $$(1+x^3+x^6+x^9+ldots)(1+x^7+x^{14}+x^{21}+ldots)$$ is exactly the number of ways for writing $n$ as $3A+7B$?
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:51








  • 1




    $begingroup$
    Well, by geometric series $(1+x^3+x^6+ldots)=frac{1}{1-x^3}$ and $(1+x^7+x^{14}+ldots)=frac{1}{1-x^7}$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:53






  • 2




    $begingroup$
    The meromorphic function $frac{1}{(1-x^3)(1-x^7)}$ has a double pole at $x=1$ but the other singularities are all simple poles. Any simple pole provides a bounded contribution, since the coefficients of $frac{1}{1-xxi}$ have unit modulus for $xiin S^1$, hence the magnitude of the representation function is linear, since $[x^n]frac{1}{(1-x)^2}=Theta(n)$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:55








  • 1




    $begingroup$
    Linear and with a bounded perturbation implies strictly positive from some point on - the hidden constant in $Theta$ is not really important, but it can be derived from stars and bars.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:56
















4












4








4





$begingroup$


I was reading about the Coin-problem and I am unable to fully understand the following argument:




On the other hand, whenever the GCD equals 1, the set of integers that cannot be expressed as a conical combination of ${ a_1, a_2, …, a_n }$ is bounded according to Schur's theorem, and therefore the Frobenius number exists.




Here the author is arguing for the existence of a non-negative solution to the linear Diophantine equation (LDE)
$$sum_{i=1}^{k}a_ix_i =n text{}$$
for large enough $n.$ Now I tried to understand the proof of the Schur theorem here enter link description here (Page 98), but I am not sure that I understand it fully. In particular, I don't understand why the generating function associated with the sequence $h_n$ that counts the number of solutions to the LDE is
$$H(x)= prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right).$$



Once we have $H(x)$ we can deduce that
$$h_nsim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$ How does this exactly show that for large enough $n,$ $h_n>0?$ Is it because the $frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}>0?$










share|cite|improve this question











$endgroup$




I was reading about the Coin-problem and I am unable to fully understand the following argument:




On the other hand, whenever the GCD equals 1, the set of integers that cannot be expressed as a conical combination of ${ a_1, a_2, …, a_n }$ is bounded according to Schur's theorem, and therefore the Frobenius number exists.




Here the author is arguing for the existence of a non-negative solution to the linear Diophantine equation (LDE)
$$sum_{i=1}^{k}a_ix_i =n text{}$$
for large enough $n.$ Now I tried to understand the proof of the Schur theorem here enter link description here (Page 98), but I am not sure that I understand it fully. In particular, I don't understand why the generating function associated with the sequence $h_n$ that counts the number of solutions to the LDE is
$$H(x)= prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right).$$



Once we have $H(x)$ we can deduce that
$$h_nsim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$ How does this exactly show that for large enough $n,$ $h_n>0?$ Is it because the $frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}>0?$







combinatorics number-theory generating-functions additive-combinatorics formal-power-series






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edited Jan 3 at 2:03









Batominovski

33.1k33293




33.1k33293










asked Jan 2 at 23:41









Hello_WorldHello_World

4,14121931




4,14121931








  • 2




    $begingroup$
    Yes.$phantom{}$
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:49






  • 3




    $begingroup$
    Consider just the case $k=2,a_1=3,a_2=7$ for clarity. Can you see that the coefficient of $x^n$ in $$(1+x^3+x^6+x^9+ldots)(1+x^7+x^{14}+x^{21}+ldots)$$ is exactly the number of ways for writing $n$ as $3A+7B$?
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:51








  • 1




    $begingroup$
    Well, by geometric series $(1+x^3+x^6+ldots)=frac{1}{1-x^3}$ and $(1+x^7+x^{14}+ldots)=frac{1}{1-x^7}$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:53






  • 2




    $begingroup$
    The meromorphic function $frac{1}{(1-x^3)(1-x^7)}$ has a double pole at $x=1$ but the other singularities are all simple poles. Any simple pole provides a bounded contribution, since the coefficients of $frac{1}{1-xxi}$ have unit modulus for $xiin S^1$, hence the magnitude of the representation function is linear, since $[x^n]frac{1}{(1-x)^2}=Theta(n)$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:55








  • 1




    $begingroup$
    Linear and with a bounded perturbation implies strictly positive from some point on - the hidden constant in $Theta$ is not really important, but it can be derived from stars and bars.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:56
















  • 2




    $begingroup$
    Yes.$phantom{}$
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:49






  • 3




    $begingroup$
    Consider just the case $k=2,a_1=3,a_2=7$ for clarity. Can you see that the coefficient of $x^n$ in $$(1+x^3+x^6+x^9+ldots)(1+x^7+x^{14}+x^{21}+ldots)$$ is exactly the number of ways for writing $n$ as $3A+7B$?
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:51








  • 1




    $begingroup$
    Well, by geometric series $(1+x^3+x^6+ldots)=frac{1}{1-x^3}$ and $(1+x^7+x^{14}+ldots)=frac{1}{1-x^7}$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:53






  • 2




    $begingroup$
    The meromorphic function $frac{1}{(1-x^3)(1-x^7)}$ has a double pole at $x=1$ but the other singularities are all simple poles. Any simple pole provides a bounded contribution, since the coefficients of $frac{1}{1-xxi}$ have unit modulus for $xiin S^1$, hence the magnitude of the representation function is linear, since $[x^n]frac{1}{(1-x)^2}=Theta(n)$.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:55








  • 1




    $begingroup$
    Linear and with a bounded perturbation implies strictly positive from some point on - the hidden constant in $Theta$ is not really important, but it can be derived from stars and bars.
    $endgroup$
    – Jack D'Aurizio
    Jan 2 at 23:56










2




2




$begingroup$
Yes.$phantom{}$
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:49




$begingroup$
Yes.$phantom{}$
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:49




3




3




$begingroup$
Consider just the case $k=2,a_1=3,a_2=7$ for clarity. Can you see that the coefficient of $x^n$ in $$(1+x^3+x^6+x^9+ldots)(1+x^7+x^{14}+x^{21}+ldots)$$ is exactly the number of ways for writing $n$ as $3A+7B$?
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:51






$begingroup$
Consider just the case $k=2,a_1=3,a_2=7$ for clarity. Can you see that the coefficient of $x^n$ in $$(1+x^3+x^6+x^9+ldots)(1+x^7+x^{14}+x^{21}+ldots)$$ is exactly the number of ways for writing $n$ as $3A+7B$?
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:51






1




1




$begingroup$
Well, by geometric series $(1+x^3+x^6+ldots)=frac{1}{1-x^3}$ and $(1+x^7+x^{14}+ldots)=frac{1}{1-x^7}$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:53




$begingroup$
Well, by geometric series $(1+x^3+x^6+ldots)=frac{1}{1-x^3}$ and $(1+x^7+x^{14}+ldots)=frac{1}{1-x^7}$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:53




2




2




$begingroup$
The meromorphic function $frac{1}{(1-x^3)(1-x^7)}$ has a double pole at $x=1$ but the other singularities are all simple poles. Any simple pole provides a bounded contribution, since the coefficients of $frac{1}{1-xxi}$ have unit modulus for $xiin S^1$, hence the magnitude of the representation function is linear, since $[x^n]frac{1}{(1-x)^2}=Theta(n)$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:55






$begingroup$
The meromorphic function $frac{1}{(1-x^3)(1-x^7)}$ has a double pole at $x=1$ but the other singularities are all simple poles. Any simple pole provides a bounded contribution, since the coefficients of $frac{1}{1-xxi}$ have unit modulus for $xiin S^1$, hence the magnitude of the representation function is linear, since $[x^n]frac{1}{(1-x)^2}=Theta(n)$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:55






1




1




$begingroup$
Linear and with a bounded perturbation implies strictly positive from some point on - the hidden constant in $Theta$ is not really important, but it can be derived from stars and bars.
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:56






$begingroup$
Linear and with a bounded perturbation implies strictly positive from some point on - the hidden constant in $Theta$ is not really important, but it can be derived from stars and bars.
$endgroup$
– Jack D'Aurizio
Jan 2 at 23:56












2 Answers
2






active

oldest

votes


















0












$begingroup$

A simple approach to the title:



Thanks to Bezout, there is some (not necessarily) integer combination of the $a_i$s that makes $1$. Add a large enough multiple of $a_1$ to each coefficient to make it non-negative and call the sum $k$; we have $kequiv 1 pmod{a_1}$.



Now any number $ge ka_1$ is a non-negative combination of the $a_i$s. Namely, let $n$ be the desired number, and let $m=nbmod a_1$. Then $mk$ is a positive combination, and we can add some multiple of $a_1$ to make it $n$ instead, since $mkequiv npmod{a_1}$ and $mk<n$ because $m<a_1$.






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  • $begingroup$
    How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
    $endgroup$
    – Mike
    Jan 4 at 18:08










  • $begingroup$
    *nonegative rational
    $endgroup$
    – Mike
    Jan 4 at 18:16






  • 1




    $begingroup$
    @Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:45












  • $begingroup$
    Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
    $endgroup$
    – Mike
    Jan 4 at 18:48












  • $begingroup$
    @Mike: Your "direct way" seems to look a lot longer than mine, though.
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:52



















0












$begingroup$

We first do this for $k =2$.



LEt us assume WLOG that $a_1 < a_2$. If gcd$(a_1,a_2)=1$ then $a_2 mod a_1 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$. So there exists a multiplicative inverse $m_2 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$ of $a_2$; or equivalently, there exists a nonegative integer $m_2 < a_1$ such that $m_2 times a_2 equiv_{a_1} 1$. This implies the following: There exists integers $m_2$ and $m_1$; $|m_2| < a_1$ and so $|m_1| leq a_2+1$ such that $m_2a_2 = 1 + m_1a_1$, or equivalently, $m_2a_2-m_1a_1 = 1$.



So let $y$ be a sufficiently large integer, and let $k in {0,1,ldots, a_1-1}$ be such that $Na_1 + k = y$. Then $Na_1 +k(m_2a_2-m_1a_1) = y$, and
$N-km_1$ is positive if $N > km_1$, where $k < a_1$ and $m_1 leq a_2+1$. This will hold if $y$ is at least $a_1(a_1+1)(a_2+1)$. This completes the case for $k=2$.



Having established this for $k=2$ we now use induction to show this holds for general $k$. In particular, let ${a_1,a_2,ldots, a_{k+1}}$ be a set of positive integers where the greatest common divisor is 1. We show using induction on $k$ that every integer $Y$ can be expressed $Y = sum_{i=1}^{k+1} M_ia_i$. To this end, let $c =$gcd$(a_k,a_{k+1})$. Then by induction, for any sufficiently large integer $y$, there are nonegative integers $M_k$ and $M_{k+1}$ such that $M_ka_k+M_{k+1}a_{k+1} = cy$. So pick $y$ sufficiently large such that gcd$(a_1,ldots, a_{k-1}, cy) = 1$; indeed any $y$ sufficiently large satisfying gcd$(a_1,ldots, a_{k-1}, y) = 1$ will do [make sure you see why], and there is indeed such a positive integer $y$ [make sure you see why].



Then by the inductive hypothesis for any sufficiently large integer $Y$ there exist nonegative integers $M_1,ldots, M_{k-1}, M_y$ such that
$cyM_y + sum_{i=1}^{k-1} M_ia_i = Y$, as $cy$ itself satisfies $cy = M_ka_k+M_{k+1}a_{k+1}$ the result follows.





ETA: HOWEVER, we do note that, setting $A = max {a_1,ldots, a_k}$ that we may assume WLOG that $k leq 1+log A$. Or more precisely, there is a subset $S$ of ${1,ldots, k}$ of cardinality $log A$ such that gcd${a_i; i in S}$ is 1. Then it would suffice to show that every significantly large number $Y$ can be written $Y=sum_{i in S} M_ia_i$ where each $M_i$ is a nonnegative integer.



Indeed, each $a_i$ can be written as a product of at most $log A$ primes. So we build a set $a_{i_1},a_{i_2},a_{i_l}; l leq 1+log A$ such that gcd${a_{i_1},ldots, a_{i_l}} = 1$. Let us assume that using induction, that gcd${a_{i_1},ldots, a_{i_j}}$ is a composite number that has at most $log A -j+1$ prime factors $p_1ldots p_{log A-j}$. Then there exists an $a_i$ such that at least one of $p_1,ldots, p_{log A-j+1}$ doesn't divide $a_i$. Then gcd${a_{i_1},ldots, a_{i_j}, a_i}$ is a composite number that has at most $log A -j$ prime factors.






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  • $begingroup$
    Editted for clarity. Caught a crucial typo!
    $endgroup$
    – Mike
    Jan 4 at 18:38











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2 Answers
2






active

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votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

A simple approach to the title:



Thanks to Bezout, there is some (not necessarily) integer combination of the $a_i$s that makes $1$. Add a large enough multiple of $a_1$ to each coefficient to make it non-negative and call the sum $k$; we have $kequiv 1 pmod{a_1}$.



Now any number $ge ka_1$ is a non-negative combination of the $a_i$s. Namely, let $n$ be the desired number, and let $m=nbmod a_1$. Then $mk$ is a positive combination, and we can add some multiple of $a_1$ to make it $n$ instead, since $mkequiv npmod{a_1}$ and $mk<n$ because $m<a_1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
    $endgroup$
    – Mike
    Jan 4 at 18:08










  • $begingroup$
    *nonegative rational
    $endgroup$
    – Mike
    Jan 4 at 18:16






  • 1




    $begingroup$
    @Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:45












  • $begingroup$
    Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
    $endgroup$
    – Mike
    Jan 4 at 18:48












  • $begingroup$
    @Mike: Your "direct way" seems to look a lot longer than mine, though.
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:52
















0












$begingroup$

A simple approach to the title:



Thanks to Bezout, there is some (not necessarily) integer combination of the $a_i$s that makes $1$. Add a large enough multiple of $a_1$ to each coefficient to make it non-negative and call the sum $k$; we have $kequiv 1 pmod{a_1}$.



Now any number $ge ka_1$ is a non-negative combination of the $a_i$s. Namely, let $n$ be the desired number, and let $m=nbmod a_1$. Then $mk$ is a positive combination, and we can add some multiple of $a_1$ to make it $n$ instead, since $mkequiv npmod{a_1}$ and $mk<n$ because $m<a_1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
    $endgroup$
    – Mike
    Jan 4 at 18:08










  • $begingroup$
    *nonegative rational
    $endgroup$
    – Mike
    Jan 4 at 18:16






  • 1




    $begingroup$
    @Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:45












  • $begingroup$
    Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
    $endgroup$
    – Mike
    Jan 4 at 18:48












  • $begingroup$
    @Mike: Your "direct way" seems to look a lot longer than mine, though.
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:52














0












0








0





$begingroup$

A simple approach to the title:



Thanks to Bezout, there is some (not necessarily) integer combination of the $a_i$s that makes $1$. Add a large enough multiple of $a_1$ to each coefficient to make it non-negative and call the sum $k$; we have $kequiv 1 pmod{a_1}$.



Now any number $ge ka_1$ is a non-negative combination of the $a_i$s. Namely, let $n$ be the desired number, and let $m=nbmod a_1$. Then $mk$ is a positive combination, and we can add some multiple of $a_1$ to make it $n$ instead, since $mkequiv npmod{a_1}$ and $mk<n$ because $m<a_1$.






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$endgroup$



A simple approach to the title:



Thanks to Bezout, there is some (not necessarily) integer combination of the $a_i$s that makes $1$. Add a large enough multiple of $a_1$ to each coefficient to make it non-negative and call the sum $k$; we have $kequiv 1 pmod{a_1}$.



Now any number $ge ka_1$ is a non-negative combination of the $a_i$s. Namely, let $n$ be the desired number, and let $m=nbmod a_1$. Then $mk$ is a positive combination, and we can add some multiple of $a_1$ to make it $n$ instead, since $mkequiv npmod{a_1}$ and $mk<n$ because $m<a_1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 4:27

























answered Jan 4 at 4:09









Henning MakholmHenning Makholm

241k17308546




241k17308546












  • $begingroup$
    How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
    $endgroup$
    – Mike
    Jan 4 at 18:08










  • $begingroup$
    *nonegative rational
    $endgroup$
    – Mike
    Jan 4 at 18:16






  • 1




    $begingroup$
    @Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:45












  • $begingroup$
    Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
    $endgroup$
    – Mike
    Jan 4 at 18:48












  • $begingroup$
    @Mike: Your "direct way" seems to look a lot longer than mine, though.
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:52


















  • $begingroup$
    How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
    $endgroup$
    – Mike
    Jan 4 at 18:08










  • $begingroup$
    *nonegative rational
    $endgroup$
    – Mike
    Jan 4 at 18:16






  • 1




    $begingroup$
    @Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:45












  • $begingroup$
    Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
    $endgroup$
    – Mike
    Jan 4 at 18:48












  • $begingroup$
    @Mike: Your "direct way" seems to look a lot longer than mine, though.
    $endgroup$
    – Henning Makholm
    Jan 4 at 18:52
















$begingroup$
How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
$endgroup$
– Mike
Jan 4 at 18:08




$begingroup$
How do you get the coefficients all integers though? I see that $k=sum c_ia_i$ where each $c_i$ is rational....
$endgroup$
– Mike
Jan 4 at 18:08












$begingroup$
*nonegative rational
$endgroup$
– Mike
Jan 4 at 18:16




$begingroup$
*nonegative rational
$endgroup$
– Mike
Jan 4 at 18:16




1




1




$begingroup$
@Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
$endgroup$
– Henning Makholm
Jan 4 at 18:45






$begingroup$
@Mike: It's a generalization of Bézout's identity. (How would they become non-integral?)
$endgroup$
– Henning Makholm
Jan 4 at 18:45














$begingroup$
Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
$endgroup$
– Mike
Jan 4 at 18:48






$begingroup$
Ahh. Gotcha. That makes sense. BUT, there is a direct way to show all this though, without having to refer to Bezout's indentity--my answer above. [I did just edit it for clarity ]
$endgroup$
– Mike
Jan 4 at 18:48














$begingroup$
@Mike: Your "direct way" seems to look a lot longer than mine, though.
$endgroup$
– Henning Makholm
Jan 4 at 18:52




$begingroup$
@Mike: Your "direct way" seems to look a lot longer than mine, though.
$endgroup$
– Henning Makholm
Jan 4 at 18:52











0












$begingroup$

We first do this for $k =2$.



LEt us assume WLOG that $a_1 < a_2$. If gcd$(a_1,a_2)=1$ then $a_2 mod a_1 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$. So there exists a multiplicative inverse $m_2 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$ of $a_2$; or equivalently, there exists a nonegative integer $m_2 < a_1$ such that $m_2 times a_2 equiv_{a_1} 1$. This implies the following: There exists integers $m_2$ and $m_1$; $|m_2| < a_1$ and so $|m_1| leq a_2+1$ such that $m_2a_2 = 1 + m_1a_1$, or equivalently, $m_2a_2-m_1a_1 = 1$.



So let $y$ be a sufficiently large integer, and let $k in {0,1,ldots, a_1-1}$ be such that $Na_1 + k = y$. Then $Na_1 +k(m_2a_2-m_1a_1) = y$, and
$N-km_1$ is positive if $N > km_1$, where $k < a_1$ and $m_1 leq a_2+1$. This will hold if $y$ is at least $a_1(a_1+1)(a_2+1)$. This completes the case for $k=2$.



Having established this for $k=2$ we now use induction to show this holds for general $k$. In particular, let ${a_1,a_2,ldots, a_{k+1}}$ be a set of positive integers where the greatest common divisor is 1. We show using induction on $k$ that every integer $Y$ can be expressed $Y = sum_{i=1}^{k+1} M_ia_i$. To this end, let $c =$gcd$(a_k,a_{k+1})$. Then by induction, for any sufficiently large integer $y$, there are nonegative integers $M_k$ and $M_{k+1}$ such that $M_ka_k+M_{k+1}a_{k+1} = cy$. So pick $y$ sufficiently large such that gcd$(a_1,ldots, a_{k-1}, cy) = 1$; indeed any $y$ sufficiently large satisfying gcd$(a_1,ldots, a_{k-1}, y) = 1$ will do [make sure you see why], and there is indeed such a positive integer $y$ [make sure you see why].



Then by the inductive hypothesis for any sufficiently large integer $Y$ there exist nonegative integers $M_1,ldots, M_{k-1}, M_y$ such that
$cyM_y + sum_{i=1}^{k-1} M_ia_i = Y$, as $cy$ itself satisfies $cy = M_ka_k+M_{k+1}a_{k+1}$ the result follows.





ETA: HOWEVER, we do note that, setting $A = max {a_1,ldots, a_k}$ that we may assume WLOG that $k leq 1+log A$. Or more precisely, there is a subset $S$ of ${1,ldots, k}$ of cardinality $log A$ such that gcd${a_i; i in S}$ is 1. Then it would suffice to show that every significantly large number $Y$ can be written $Y=sum_{i in S} M_ia_i$ where each $M_i$ is a nonnegative integer.



Indeed, each $a_i$ can be written as a product of at most $log A$ primes. So we build a set $a_{i_1},a_{i_2},a_{i_l}; l leq 1+log A$ such that gcd${a_{i_1},ldots, a_{i_l}} = 1$. Let us assume that using induction, that gcd${a_{i_1},ldots, a_{i_j}}$ is a composite number that has at most $log A -j+1$ prime factors $p_1ldots p_{log A-j}$. Then there exists an $a_i$ such that at least one of $p_1,ldots, p_{log A-j+1}$ doesn't divide $a_i$. Then gcd${a_{i_1},ldots, a_{i_j}, a_i}$ is a composite number that has at most $log A -j$ prime factors.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Editted for clarity. Caught a crucial typo!
    $endgroup$
    – Mike
    Jan 4 at 18:38
















0












$begingroup$

We first do this for $k =2$.



LEt us assume WLOG that $a_1 < a_2$. If gcd$(a_1,a_2)=1$ then $a_2 mod a_1 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$. So there exists a multiplicative inverse $m_2 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$ of $a_2$; or equivalently, there exists a nonegative integer $m_2 < a_1$ such that $m_2 times a_2 equiv_{a_1} 1$. This implies the following: There exists integers $m_2$ and $m_1$; $|m_2| < a_1$ and so $|m_1| leq a_2+1$ such that $m_2a_2 = 1 + m_1a_1$, or equivalently, $m_2a_2-m_1a_1 = 1$.



So let $y$ be a sufficiently large integer, and let $k in {0,1,ldots, a_1-1}$ be such that $Na_1 + k = y$. Then $Na_1 +k(m_2a_2-m_1a_1) = y$, and
$N-km_1$ is positive if $N > km_1$, where $k < a_1$ and $m_1 leq a_2+1$. This will hold if $y$ is at least $a_1(a_1+1)(a_2+1)$. This completes the case for $k=2$.



Having established this for $k=2$ we now use induction to show this holds for general $k$. In particular, let ${a_1,a_2,ldots, a_{k+1}}$ be a set of positive integers where the greatest common divisor is 1. We show using induction on $k$ that every integer $Y$ can be expressed $Y = sum_{i=1}^{k+1} M_ia_i$. To this end, let $c =$gcd$(a_k,a_{k+1})$. Then by induction, for any sufficiently large integer $y$, there are nonegative integers $M_k$ and $M_{k+1}$ such that $M_ka_k+M_{k+1}a_{k+1} = cy$. So pick $y$ sufficiently large such that gcd$(a_1,ldots, a_{k-1}, cy) = 1$; indeed any $y$ sufficiently large satisfying gcd$(a_1,ldots, a_{k-1}, y) = 1$ will do [make sure you see why], and there is indeed such a positive integer $y$ [make sure you see why].



Then by the inductive hypothesis for any sufficiently large integer $Y$ there exist nonegative integers $M_1,ldots, M_{k-1}, M_y$ such that
$cyM_y + sum_{i=1}^{k-1} M_ia_i = Y$, as $cy$ itself satisfies $cy = M_ka_k+M_{k+1}a_{k+1}$ the result follows.





ETA: HOWEVER, we do note that, setting $A = max {a_1,ldots, a_k}$ that we may assume WLOG that $k leq 1+log A$. Or more precisely, there is a subset $S$ of ${1,ldots, k}$ of cardinality $log A$ such that gcd${a_i; i in S}$ is 1. Then it would suffice to show that every significantly large number $Y$ can be written $Y=sum_{i in S} M_ia_i$ where each $M_i$ is a nonnegative integer.



Indeed, each $a_i$ can be written as a product of at most $log A$ primes. So we build a set $a_{i_1},a_{i_2},a_{i_l}; l leq 1+log A$ such that gcd${a_{i_1},ldots, a_{i_l}} = 1$. Let us assume that using induction, that gcd${a_{i_1},ldots, a_{i_j}}$ is a composite number that has at most $log A -j+1$ prime factors $p_1ldots p_{log A-j}$. Then there exists an $a_i$ such that at least one of $p_1,ldots, p_{log A-j+1}$ doesn't divide $a_i$. Then gcd${a_{i_1},ldots, a_{i_j}, a_i}$ is a composite number that has at most $log A -j$ prime factors.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Editted for clarity. Caught a crucial typo!
    $endgroup$
    – Mike
    Jan 4 at 18:38














0












0








0





$begingroup$

We first do this for $k =2$.



LEt us assume WLOG that $a_1 < a_2$. If gcd$(a_1,a_2)=1$ then $a_2 mod a_1 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$. So there exists a multiplicative inverse $m_2 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$ of $a_2$; or equivalently, there exists a nonegative integer $m_2 < a_1$ such that $m_2 times a_2 equiv_{a_1} 1$. This implies the following: There exists integers $m_2$ and $m_1$; $|m_2| < a_1$ and so $|m_1| leq a_2+1$ such that $m_2a_2 = 1 + m_1a_1$, or equivalently, $m_2a_2-m_1a_1 = 1$.



So let $y$ be a sufficiently large integer, and let $k in {0,1,ldots, a_1-1}$ be such that $Na_1 + k = y$. Then $Na_1 +k(m_2a_2-m_1a_1) = y$, and
$N-km_1$ is positive if $N > km_1$, where $k < a_1$ and $m_1 leq a_2+1$. This will hold if $y$ is at least $a_1(a_1+1)(a_2+1)$. This completes the case for $k=2$.



Having established this for $k=2$ we now use induction to show this holds for general $k$. In particular, let ${a_1,a_2,ldots, a_{k+1}}$ be a set of positive integers where the greatest common divisor is 1. We show using induction on $k$ that every integer $Y$ can be expressed $Y = sum_{i=1}^{k+1} M_ia_i$. To this end, let $c =$gcd$(a_k,a_{k+1})$. Then by induction, for any sufficiently large integer $y$, there are nonegative integers $M_k$ and $M_{k+1}$ such that $M_ka_k+M_{k+1}a_{k+1} = cy$. So pick $y$ sufficiently large such that gcd$(a_1,ldots, a_{k-1}, cy) = 1$; indeed any $y$ sufficiently large satisfying gcd$(a_1,ldots, a_{k-1}, y) = 1$ will do [make sure you see why], and there is indeed such a positive integer $y$ [make sure you see why].



Then by the inductive hypothesis for any sufficiently large integer $Y$ there exist nonegative integers $M_1,ldots, M_{k-1}, M_y$ such that
$cyM_y + sum_{i=1}^{k-1} M_ia_i = Y$, as $cy$ itself satisfies $cy = M_ka_k+M_{k+1}a_{k+1}$ the result follows.





ETA: HOWEVER, we do note that, setting $A = max {a_1,ldots, a_k}$ that we may assume WLOG that $k leq 1+log A$. Or more precisely, there is a subset $S$ of ${1,ldots, k}$ of cardinality $log A$ such that gcd${a_i; i in S}$ is 1. Then it would suffice to show that every significantly large number $Y$ can be written $Y=sum_{i in S} M_ia_i$ where each $M_i$ is a nonnegative integer.



Indeed, each $a_i$ can be written as a product of at most $log A$ primes. So we build a set $a_{i_1},a_{i_2},a_{i_l}; l leq 1+log A$ such that gcd${a_{i_1},ldots, a_{i_l}} = 1$. Let us assume that using induction, that gcd${a_{i_1},ldots, a_{i_j}}$ is a composite number that has at most $log A -j+1$ prime factors $p_1ldots p_{log A-j}$. Then there exists an $a_i$ such that at least one of $p_1,ldots, p_{log A-j+1}$ doesn't divide $a_i$. Then gcd${a_{i_1},ldots, a_{i_j}, a_i}$ is a composite number that has at most $log A -j$ prime factors.






share|cite|improve this answer











$endgroup$



We first do this for $k =2$.



LEt us assume WLOG that $a_1 < a_2$. If gcd$(a_1,a_2)=1$ then $a_2 mod a_1 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$. So there exists a multiplicative inverse $m_2 in left(mathbb{Z}/a_1 mathbb{Z}right)^{times}$ of $a_2$; or equivalently, there exists a nonegative integer $m_2 < a_1$ such that $m_2 times a_2 equiv_{a_1} 1$. This implies the following: There exists integers $m_2$ and $m_1$; $|m_2| < a_1$ and so $|m_1| leq a_2+1$ such that $m_2a_2 = 1 + m_1a_1$, or equivalently, $m_2a_2-m_1a_1 = 1$.



So let $y$ be a sufficiently large integer, and let $k in {0,1,ldots, a_1-1}$ be such that $Na_1 + k = y$. Then $Na_1 +k(m_2a_2-m_1a_1) = y$, and
$N-km_1$ is positive if $N > km_1$, where $k < a_1$ and $m_1 leq a_2+1$. This will hold if $y$ is at least $a_1(a_1+1)(a_2+1)$. This completes the case for $k=2$.



Having established this for $k=2$ we now use induction to show this holds for general $k$. In particular, let ${a_1,a_2,ldots, a_{k+1}}$ be a set of positive integers where the greatest common divisor is 1. We show using induction on $k$ that every integer $Y$ can be expressed $Y = sum_{i=1}^{k+1} M_ia_i$. To this end, let $c =$gcd$(a_k,a_{k+1})$. Then by induction, for any sufficiently large integer $y$, there are nonegative integers $M_k$ and $M_{k+1}$ such that $M_ka_k+M_{k+1}a_{k+1} = cy$. So pick $y$ sufficiently large such that gcd$(a_1,ldots, a_{k-1}, cy) = 1$; indeed any $y$ sufficiently large satisfying gcd$(a_1,ldots, a_{k-1}, y) = 1$ will do [make sure you see why], and there is indeed such a positive integer $y$ [make sure you see why].



Then by the inductive hypothesis for any sufficiently large integer $Y$ there exist nonegative integers $M_1,ldots, M_{k-1}, M_y$ such that
$cyM_y + sum_{i=1}^{k-1} M_ia_i = Y$, as $cy$ itself satisfies $cy = M_ka_k+M_{k+1}a_{k+1}$ the result follows.





ETA: HOWEVER, we do note that, setting $A = max {a_1,ldots, a_k}$ that we may assume WLOG that $k leq 1+log A$. Or more precisely, there is a subset $S$ of ${1,ldots, k}$ of cardinality $log A$ such that gcd${a_i; i in S}$ is 1. Then it would suffice to show that every significantly large number $Y$ can be written $Y=sum_{i in S} M_ia_i$ where each $M_i$ is a nonnegative integer.



Indeed, each $a_i$ can be written as a product of at most $log A$ primes. So we build a set $a_{i_1},a_{i_2},a_{i_l}; l leq 1+log A$ such that gcd${a_{i_1},ldots, a_{i_l}} = 1$. Let us assume that using induction, that gcd${a_{i_1},ldots, a_{i_j}}$ is a composite number that has at most $log A -j+1$ prime factors $p_1ldots p_{log A-j}$. Then there exists an $a_i$ such that at least one of $p_1,ldots, p_{log A-j+1}$ doesn't divide $a_i$. Then gcd${a_{i_1},ldots, a_{i_j}, a_i}$ is a composite number that has at most $log A -j$ prime factors.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 5:01

























answered Jan 4 at 0:47









MikeMike

4,266412




4,266412












  • $begingroup$
    Editted for clarity. Caught a crucial typo!
    $endgroup$
    – Mike
    Jan 4 at 18:38


















  • $begingroup$
    Editted for clarity. Caught a crucial typo!
    $endgroup$
    – Mike
    Jan 4 at 18:38
















$begingroup$
Editted for clarity. Caught a crucial typo!
$endgroup$
– Mike
Jan 4 at 18:38




$begingroup$
Editted for clarity. Caught a crucial typo!
$endgroup$
– Mike
Jan 4 at 18:38


















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