A question about the number 541456
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I found that the concatenation in base 10 of $2^{541456}-1$ and $2^{541455}-1$, gives a probable prime.
I also found about number $541456$ that:
$$5cdot(5^2+4^2+1+4^2+5^2+6^2)=(5^3+4^3+1+4^3+5^3+6^3)$$
the equation $acdot(2cdot a^2+2cdot b^2+c^2+d^2)=(2cdot a^3+2cdot b^3+c^3+d^3)$ besides $a=5, b=4, c=6, d=1$ has other non trivial solutions with a,b,c,d positive integers?
number-theory
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show 2 more comments
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I found that the concatenation in base 10 of $2^{541456}-1$ and $2^{541455}-1$, gives a probable prime.
I also found about number $541456$ that:
$$5cdot(5^2+4^2+1+4^2+5^2+6^2)=(5^3+4^3+1+4^3+5^3+6^3)$$
the equation $acdot(2cdot a^2+2cdot b^2+c^2+d^2)=(2cdot a^3+2cdot b^3+c^3+d^3)$ besides $a=5, b=4, c=6, d=1$ has other non trivial solutions with a,b,c,d positive integers?
number-theory
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Do you mean different values? If not, take $a=b=c=d=1$.
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– AugSB
Jan 4 at 20:52
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As a rough estimate based on the prime number theorem there ought to be about three $k$-digit numbers such that the base-ten concatenation of $2^{n+1}-1$ and $2^n-1$ is prime, for each $k$.
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– Henning Makholm
Jan 4 at 20:58
1
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A suggestion: replace the two $d$'s in the equation you ask about with a $d^2$ on the left and a $d^3$ on the right.
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– Barry Cipra
Jan 4 at 21:08
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Why not make this homogeneous on each side since $1=1^2=1^3$? This is equivalent to @BarryCipra's comment.
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– Mark Bennet
Jan 4 at 21:13
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@Barry Cipra now i replaced as you suggested
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– user631751
Jan 4 at 21:33
|
show 2 more comments
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I found that the concatenation in base 10 of $2^{541456}-1$ and $2^{541455}-1$, gives a probable prime.
I also found about number $541456$ that:
$$5cdot(5^2+4^2+1+4^2+5^2+6^2)=(5^3+4^3+1+4^3+5^3+6^3)$$
the equation $acdot(2cdot a^2+2cdot b^2+c^2+d^2)=(2cdot a^3+2cdot b^3+c^3+d^3)$ besides $a=5, b=4, c=6, d=1$ has other non trivial solutions with a,b,c,d positive integers?
number-theory
$endgroup$
I found that the concatenation in base 10 of $2^{541456}-1$ and $2^{541455}-1$, gives a probable prime.
I also found about number $541456$ that:
$$5cdot(5^2+4^2+1+4^2+5^2+6^2)=(5^3+4^3+1+4^3+5^3+6^3)$$
the equation $acdot(2cdot a^2+2cdot b^2+c^2+d^2)=(2cdot a^3+2cdot b^3+c^3+d^3)$ besides $a=5, b=4, c=6, d=1$ has other non trivial solutions with a,b,c,d positive integers?
number-theory
number-theory
edited Jan 4 at 21:20
user631751
asked Jan 4 at 20:39
user631751user631751
62
62
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Do you mean different values? If not, take $a=b=c=d=1$.
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– AugSB
Jan 4 at 20:52
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As a rough estimate based on the prime number theorem there ought to be about three $k$-digit numbers such that the base-ten concatenation of $2^{n+1}-1$ and $2^n-1$ is prime, for each $k$.
$endgroup$
– Henning Makholm
Jan 4 at 20:58
1
$begingroup$
A suggestion: replace the two $d$'s in the equation you ask about with a $d^2$ on the left and a $d^3$ on the right.
$endgroup$
– Barry Cipra
Jan 4 at 21:08
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Why not make this homogeneous on each side since $1=1^2=1^3$? This is equivalent to @BarryCipra's comment.
$endgroup$
– Mark Bennet
Jan 4 at 21:13
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@Barry Cipra now i replaced as you suggested
$endgroup$
– user631751
Jan 4 at 21:33
|
show 2 more comments
$begingroup$
Do you mean different values? If not, take $a=b=c=d=1$.
$endgroup$
– AugSB
Jan 4 at 20:52
$begingroup$
As a rough estimate based on the prime number theorem there ought to be about three $k$-digit numbers such that the base-ten concatenation of $2^{n+1}-1$ and $2^n-1$ is prime, for each $k$.
$endgroup$
– Henning Makholm
Jan 4 at 20:58
1
$begingroup$
A suggestion: replace the two $d$'s in the equation you ask about with a $d^2$ on the left and a $d^3$ on the right.
$endgroup$
– Barry Cipra
Jan 4 at 21:08
$begingroup$
Why not make this homogeneous on each side since $1=1^2=1^3$? This is equivalent to @BarryCipra's comment.
$endgroup$
– Mark Bennet
Jan 4 at 21:13
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@Barry Cipra now i replaced as you suggested
$endgroup$
– user631751
Jan 4 at 21:33
$begingroup$
Do you mean different values? If not, take $a=b=c=d=1$.
$endgroup$
– AugSB
Jan 4 at 20:52
$begingroup$
Do you mean different values? If not, take $a=b=c=d=1$.
$endgroup$
– AugSB
Jan 4 at 20:52
$begingroup$
As a rough estimate based on the prime number theorem there ought to be about three $k$-digit numbers such that the base-ten concatenation of $2^{n+1}-1$ and $2^n-1$ is prime, for each $k$.
$endgroup$
– Henning Makholm
Jan 4 at 20:58
$begingroup$
As a rough estimate based on the prime number theorem there ought to be about three $k$-digit numbers such that the base-ten concatenation of $2^{n+1}-1$ and $2^n-1$ is prime, for each $k$.
$endgroup$
– Henning Makholm
Jan 4 at 20:58
1
1
$begingroup$
A suggestion: replace the two $d$'s in the equation you ask about with a $d^2$ on the left and a $d^3$ on the right.
$endgroup$
– Barry Cipra
Jan 4 at 21:08
$begingroup$
A suggestion: replace the two $d$'s in the equation you ask about with a $d^2$ on the left and a $d^3$ on the right.
$endgroup$
– Barry Cipra
Jan 4 at 21:08
$begingroup$
Why not make this homogeneous on each side since $1=1^2=1^3$? This is equivalent to @BarryCipra's comment.
$endgroup$
– Mark Bennet
Jan 4 at 21:13
$begingroup$
Why not make this homogeneous on each side since $1=1^2=1^3$? This is equivalent to @BarryCipra's comment.
$endgroup$
– Mark Bennet
Jan 4 at 21:13
$begingroup$
@Barry Cipra now i replaced as you suggested
$endgroup$
– user631751
Jan 4 at 21:33
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@Barry Cipra now i replaced as you suggested
$endgroup$
– user631751
Jan 4 at 21:33
|
show 2 more comments
3 Answers
3
active
oldest
votes
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Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.
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Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
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– user631751
Jan 4 at 22:05
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That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
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– Ross Millikan
Jan 4 at 22:12
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can you proof that?
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– user631751
Jan 4 at 22:14
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I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
add a comment |
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This section of this Answer is responsive to the version of the Question which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d^2) = (2 a^3 + 2 b^3 + c^3 + d^3)$.
Notice that $c$ and $d$ are interchangeable, so we may impose $c leq d$.
One way to look at this is $b,c,d geq 1$ and $a =frac{2 b^3 + c^3 + d^3}{2 b^2 + c^2 + d^2}$, when that is an integer. This leads to the question, does that ever happen? Yes. Notice if $b = c = d$, this is $a = (4b^3)/(4 b^2) = b$, so $a = b = c = d$ is a solution for any choice of $a$.
begin{align*}
&(a,b,c,d) & text{giving} \
&(4,4,4,4) & 384 \
&(5,5,5,5) & 750 \
end{align*}
Cylindrical algebraic decomposition with variable order $c, d, a, b$, gives five cylindrical components. Exhibiting an integer point in a component or showing that a component has no integer points is not easy.
$c geq 1$, $d geq 1$, $a = b = frac{c^3 + d^3}{c^2 + d^2}$, when this is an integer. The fraction is always an integer when $c = d$ and is sometimes an integer otherwise. Examples: $(1,1,1,1)$, $(9,9,5,10)$
$d geq c geq 1$, $a > frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the least real root of $2x^3 -2ax^2 -a(c^2 +d^2) +c^3 + d^3$. Examples: $(9,10,5,5)$, $(17,18,2,14)$, and $(251,255,51,251)$.
$d geq c geq 1$, $a$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when those roots are integers. Having exhausted my tools for determining whether this components has integral points, I am unable to find any or prove that there are none. (Taking intersections with CADs having different variable orders splits this component into pieces, for instance $a geq 2$, $b = 2 a/3$, $c = 2 a/3$, and $d$ is the least real root of $9 x^3 - 9 a x^2 - 4 a^3 = 0$. It's unclear whether that root is ever an integer for an integer $a$ divisible by $3$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(4,2,3,5)$ and $(5,2,2,6)$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the largest root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(5,4,1,6)$ and $(17,15,5,20)$.
The below is responsive to the first version of the Question, which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d) = (2 a^3 + 2 b^3 + c^3 + d)$.
Well, $a = b = c = d = 1$ is pretty obvious, giving $6$ on both sides.
One infinite family is $a=b=c = 1$, $d$ any positive integer, giving $d+5$. There don't seem to be more with $a = 1$, but I don't know how to prove that.
Here are eight more.
begin{align*}
&(a,b,c,d) & text{giving} \
&(2, 1, 7, 243) & 604 \
&(2, 1, 100, 979998) & 1980016 \
&(2, 2, 8, 384) & 928 \
&(2, 2, 68, 305184) & 619648 \
&(2, 3, 19, 6155) & 13084 \
&(2, 41, 1, 131117) & 268976 \
&(2, 41, 26, 147342) & 302776 \
&(2, 41, 27, 149343) & 306884
end{align*}
Cylindrical algebraic decomposition, using the variable ordering $b,c,a,d$ of your equation gives three solution families:
$a = b = c = 1$ and $d geq 1$, e.g., $(1,1,1,1)$ and $(1,1,1,108)$.
$b = 1$, $c geq 2$, $1 < a < frac{c^3+2}{c^2+2}$, $d = frac{c^3 - a c^2 - 2a + 2}{a-1}$, when $d$ is an integer, e.g., $(2,1,3,7)$ and $(109,1,126,2497)$.
$b geq 2$, $c geq 1$, $1 < a < frac{c^3+2b^3}{c^2+2b^2}$, $d = frac{c^4 - a c^2 + 2 b^3-2ab^2}{a-1}$, when $d$ is an integer, e.g., $(5,2,6,3)$ and $(3,3,100,485,000)$.
(The variable ordering $a,b,c,d$ gives more families, some of which appear to be empty, and most of which require bounds on $b$ and $c$ that are somewhat complicated algebraic functions of the prior variables. The previous version(s) of this answer used that ordering, but the new ordering is used to aid readability.)
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what if $aneq bneq cneq d$?
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– Enzo Creti
Jan 5 at 19:12
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they should be infinite
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– Enzo Creti
Jan 5 at 20:37
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no no ok thank you!
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– Enzo Creti
Jan 5 at 20:40
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@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
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– Eric Towers
Jan 6 at 7:04
add a comment |
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An obvious solution lies where a,b,c,d all equal 1 as:
$$1cdot(2cdot1^2+2cdot1^2+1^2+1) = (2cdot1^3+2cdot1^3+1^3+1)$$
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this seems a trivial solution
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– user631751
Jan 4 at 20:58
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.
$endgroup$
$begingroup$
Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
$endgroup$
– user631751
Jan 4 at 22:05
$begingroup$
That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
$endgroup$
– Ross Millikan
Jan 4 at 22:12
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can you proof that?
$endgroup$
– user631751
Jan 4 at 22:14
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I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
add a comment |
$begingroup$
Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.
$endgroup$
$begingroup$
Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
$endgroup$
– user631751
Jan 4 at 22:05
$begingroup$
That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
$endgroup$
– Ross Millikan
Jan 4 at 22:12
$begingroup$
can you proof that?
$endgroup$
– user631751
Jan 4 at 22:14
$begingroup$
I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
add a comment |
$begingroup$
Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.
$endgroup$
Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.
edited Jan 4 at 21:15
answered Jan 4 at 20:59
Ross MillikanRoss Millikan
298k24200373
298k24200373
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Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
$endgroup$
– user631751
Jan 4 at 22:05
$begingroup$
That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
$endgroup$
– Ross Millikan
Jan 4 at 22:12
$begingroup$
can you proof that?
$endgroup$
– user631751
Jan 4 at 22:14
$begingroup$
I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
add a comment |
$begingroup$
Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
$endgroup$
– user631751
Jan 4 at 22:05
$begingroup$
That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
$endgroup$
– Ross Millikan
Jan 4 at 22:12
$begingroup$
can you proof that?
$endgroup$
– user631751
Jan 4 at 22:14
$begingroup$
I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
$begingroup$
Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
$endgroup$
– user631751
Jan 4 at 22:05
$begingroup$
Mllikan besides a=1, b=1, c=1, d=1, the only solution with d=1 seems to be a=5,b=4, c=6, d=1?
$endgroup$
– user631751
Jan 4 at 22:05
$begingroup$
That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
$endgroup$
– Ross Millikan
Jan 4 at 22:12
$begingroup$
That is correct if you want $a,b,c,d$ to be nonzero digits. If you don't restrict the values to $9$ or below Eric Towers has more cases.
$endgroup$
– Ross Millikan
Jan 4 at 22:12
$begingroup$
can you proof that?
$endgroup$
– user631751
Jan 4 at 22:14
$begingroup$
can you proof that?
$endgroup$
– user631751
Jan 4 at 22:14
$begingroup$
I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
$begingroup$
I did a complete search in the space I described. That is a fine proof (assuming it is done correctly). There are only $9^4=6561$ cases to check
$endgroup$
– Ross Millikan
Jan 4 at 22:16
add a comment |
$begingroup$
This section of this Answer is responsive to the version of the Question which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d^2) = (2 a^3 + 2 b^3 + c^3 + d^3)$.
Notice that $c$ and $d$ are interchangeable, so we may impose $c leq d$.
One way to look at this is $b,c,d geq 1$ and $a =frac{2 b^3 + c^3 + d^3}{2 b^2 + c^2 + d^2}$, when that is an integer. This leads to the question, does that ever happen? Yes. Notice if $b = c = d$, this is $a = (4b^3)/(4 b^2) = b$, so $a = b = c = d$ is a solution for any choice of $a$.
begin{align*}
&(a,b,c,d) & text{giving} \
&(4,4,4,4) & 384 \
&(5,5,5,5) & 750 \
end{align*}
Cylindrical algebraic decomposition with variable order $c, d, a, b$, gives five cylindrical components. Exhibiting an integer point in a component or showing that a component has no integer points is not easy.
$c geq 1$, $d geq 1$, $a = b = frac{c^3 + d^3}{c^2 + d^2}$, when this is an integer. The fraction is always an integer when $c = d$ and is sometimes an integer otherwise. Examples: $(1,1,1,1)$, $(9,9,5,10)$
$d geq c geq 1$, $a > frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the least real root of $2x^3 -2ax^2 -a(c^2 +d^2) +c^3 + d^3$. Examples: $(9,10,5,5)$, $(17,18,2,14)$, and $(251,255,51,251)$.
$d geq c geq 1$, $a$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when those roots are integers. Having exhausted my tools for determining whether this components has integral points, I am unable to find any or prove that there are none. (Taking intersections with CADs having different variable orders splits this component into pieces, for instance $a geq 2$, $b = 2 a/3$, $c = 2 a/3$, and $d$ is the least real root of $9 x^3 - 9 a x^2 - 4 a^3 = 0$. It's unclear whether that root is ever an integer for an integer $a$ divisible by $3$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(4,2,3,5)$ and $(5,2,2,6)$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the largest root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(5,4,1,6)$ and $(17,15,5,20)$.
The below is responsive to the first version of the Question, which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d) = (2 a^3 + 2 b^3 + c^3 + d)$.
Well, $a = b = c = d = 1$ is pretty obvious, giving $6$ on both sides.
One infinite family is $a=b=c = 1$, $d$ any positive integer, giving $d+5$. There don't seem to be more with $a = 1$, but I don't know how to prove that.
Here are eight more.
begin{align*}
&(a,b,c,d) & text{giving} \
&(2, 1, 7, 243) & 604 \
&(2, 1, 100, 979998) & 1980016 \
&(2, 2, 8, 384) & 928 \
&(2, 2, 68, 305184) & 619648 \
&(2, 3, 19, 6155) & 13084 \
&(2, 41, 1, 131117) & 268976 \
&(2, 41, 26, 147342) & 302776 \
&(2, 41, 27, 149343) & 306884
end{align*}
Cylindrical algebraic decomposition, using the variable ordering $b,c,a,d$ of your equation gives three solution families:
$a = b = c = 1$ and $d geq 1$, e.g., $(1,1,1,1)$ and $(1,1,1,108)$.
$b = 1$, $c geq 2$, $1 < a < frac{c^3+2}{c^2+2}$, $d = frac{c^3 - a c^2 - 2a + 2}{a-1}$, when $d$ is an integer, e.g., $(2,1,3,7)$ and $(109,1,126,2497)$.
$b geq 2$, $c geq 1$, $1 < a < frac{c^3+2b^3}{c^2+2b^2}$, $d = frac{c^4 - a c^2 + 2 b^3-2ab^2}{a-1}$, when $d$ is an integer, e.g., $(5,2,6,3)$ and $(3,3,100,485,000)$.
(The variable ordering $a,b,c,d$ gives more families, some of which appear to be empty, and most of which require bounds on $b$ and $c$ that are somewhat complicated algebraic functions of the prior variables. The previous version(s) of this answer used that ordering, but the new ordering is used to aid readability.)
$endgroup$
$begingroup$
what if $aneq bneq cneq d$?
$endgroup$
– Enzo Creti
Jan 5 at 19:12
$begingroup$
they should be infinite
$endgroup$
– Enzo Creti
Jan 5 at 20:37
$begingroup$
no no ok thank you!
$endgroup$
– Enzo Creti
Jan 5 at 20:40
$begingroup$
@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
$endgroup$
– Eric Towers
Jan 6 at 7:04
add a comment |
$begingroup$
This section of this Answer is responsive to the version of the Question which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d^2) = (2 a^3 + 2 b^3 + c^3 + d^3)$.
Notice that $c$ and $d$ are interchangeable, so we may impose $c leq d$.
One way to look at this is $b,c,d geq 1$ and $a =frac{2 b^3 + c^3 + d^3}{2 b^2 + c^2 + d^2}$, when that is an integer. This leads to the question, does that ever happen? Yes. Notice if $b = c = d$, this is $a = (4b^3)/(4 b^2) = b$, so $a = b = c = d$ is a solution for any choice of $a$.
begin{align*}
&(a,b,c,d) & text{giving} \
&(4,4,4,4) & 384 \
&(5,5,5,5) & 750 \
end{align*}
Cylindrical algebraic decomposition with variable order $c, d, a, b$, gives five cylindrical components. Exhibiting an integer point in a component or showing that a component has no integer points is not easy.
$c geq 1$, $d geq 1$, $a = b = frac{c^3 + d^3}{c^2 + d^2}$, when this is an integer. The fraction is always an integer when $c = d$ and is sometimes an integer otherwise. Examples: $(1,1,1,1)$, $(9,9,5,10)$
$d geq c geq 1$, $a > frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the least real root of $2x^3 -2ax^2 -a(c^2 +d^2) +c^3 + d^3$. Examples: $(9,10,5,5)$, $(17,18,2,14)$, and $(251,255,51,251)$.
$d geq c geq 1$, $a$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when those roots are integers. Having exhausted my tools for determining whether this components has integral points, I am unable to find any or prove that there are none. (Taking intersections with CADs having different variable orders splits this component into pieces, for instance $a geq 2$, $b = 2 a/3$, $c = 2 a/3$, and $d$ is the least real root of $9 x^3 - 9 a x^2 - 4 a^3 = 0$. It's unclear whether that root is ever an integer for an integer $a$ divisible by $3$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(4,2,3,5)$ and $(5,2,2,6)$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the largest root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(5,4,1,6)$ and $(17,15,5,20)$.
The below is responsive to the first version of the Question, which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d) = (2 a^3 + 2 b^3 + c^3 + d)$.
Well, $a = b = c = d = 1$ is pretty obvious, giving $6$ on both sides.
One infinite family is $a=b=c = 1$, $d$ any positive integer, giving $d+5$. There don't seem to be more with $a = 1$, but I don't know how to prove that.
Here are eight more.
begin{align*}
&(a,b,c,d) & text{giving} \
&(2, 1, 7, 243) & 604 \
&(2, 1, 100, 979998) & 1980016 \
&(2, 2, 8, 384) & 928 \
&(2, 2, 68, 305184) & 619648 \
&(2, 3, 19, 6155) & 13084 \
&(2, 41, 1, 131117) & 268976 \
&(2, 41, 26, 147342) & 302776 \
&(2, 41, 27, 149343) & 306884
end{align*}
Cylindrical algebraic decomposition, using the variable ordering $b,c,a,d$ of your equation gives three solution families:
$a = b = c = 1$ and $d geq 1$, e.g., $(1,1,1,1)$ and $(1,1,1,108)$.
$b = 1$, $c geq 2$, $1 < a < frac{c^3+2}{c^2+2}$, $d = frac{c^3 - a c^2 - 2a + 2}{a-1}$, when $d$ is an integer, e.g., $(2,1,3,7)$ and $(109,1,126,2497)$.
$b geq 2$, $c geq 1$, $1 < a < frac{c^3+2b^3}{c^2+2b^2}$, $d = frac{c^4 - a c^2 + 2 b^3-2ab^2}{a-1}$, when $d$ is an integer, e.g., $(5,2,6,3)$ and $(3,3,100,485,000)$.
(The variable ordering $a,b,c,d$ gives more families, some of which appear to be empty, and most of which require bounds on $b$ and $c$ that are somewhat complicated algebraic functions of the prior variables. The previous version(s) of this answer used that ordering, but the new ordering is used to aid readability.)
$endgroup$
$begingroup$
what if $aneq bneq cneq d$?
$endgroup$
– Enzo Creti
Jan 5 at 19:12
$begingroup$
they should be infinite
$endgroup$
– Enzo Creti
Jan 5 at 20:37
$begingroup$
no no ok thank you!
$endgroup$
– Enzo Creti
Jan 5 at 20:40
$begingroup$
@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
$endgroup$
– Eric Towers
Jan 6 at 7:04
add a comment |
$begingroup$
This section of this Answer is responsive to the version of the Question which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d^2) = (2 a^3 + 2 b^3 + c^3 + d^3)$.
Notice that $c$ and $d$ are interchangeable, so we may impose $c leq d$.
One way to look at this is $b,c,d geq 1$ and $a =frac{2 b^3 + c^3 + d^3}{2 b^2 + c^2 + d^2}$, when that is an integer. This leads to the question, does that ever happen? Yes. Notice if $b = c = d$, this is $a = (4b^3)/(4 b^2) = b$, so $a = b = c = d$ is a solution for any choice of $a$.
begin{align*}
&(a,b,c,d) & text{giving} \
&(4,4,4,4) & 384 \
&(5,5,5,5) & 750 \
end{align*}
Cylindrical algebraic decomposition with variable order $c, d, a, b$, gives five cylindrical components. Exhibiting an integer point in a component or showing that a component has no integer points is not easy.
$c geq 1$, $d geq 1$, $a = b = frac{c^3 + d^3}{c^2 + d^2}$, when this is an integer. The fraction is always an integer when $c = d$ and is sometimes an integer otherwise. Examples: $(1,1,1,1)$, $(9,9,5,10)$
$d geq c geq 1$, $a > frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the least real root of $2x^3 -2ax^2 -a(c^2 +d^2) +c^3 + d^3$. Examples: $(9,10,5,5)$, $(17,18,2,14)$, and $(251,255,51,251)$.
$d geq c geq 1$, $a$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when those roots are integers. Having exhausted my tools for determining whether this components has integral points, I am unable to find any or prove that there are none. (Taking intersections with CADs having different variable orders splits this component into pieces, for instance $a geq 2$, $b = 2 a/3$, $c = 2 a/3$, and $d$ is the least real root of $9 x^3 - 9 a x^2 - 4 a^3 = 0$. It's unclear whether that root is ever an integer for an integer $a$ divisible by $3$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(4,2,3,5)$ and $(5,2,2,6)$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the largest root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(5,4,1,6)$ and $(17,15,5,20)$.
The below is responsive to the first version of the Question, which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d) = (2 a^3 + 2 b^3 + c^3 + d)$.
Well, $a = b = c = d = 1$ is pretty obvious, giving $6$ on both sides.
One infinite family is $a=b=c = 1$, $d$ any positive integer, giving $d+5$. There don't seem to be more with $a = 1$, but I don't know how to prove that.
Here are eight more.
begin{align*}
&(a,b,c,d) & text{giving} \
&(2, 1, 7, 243) & 604 \
&(2, 1, 100, 979998) & 1980016 \
&(2, 2, 8, 384) & 928 \
&(2, 2, 68, 305184) & 619648 \
&(2, 3, 19, 6155) & 13084 \
&(2, 41, 1, 131117) & 268976 \
&(2, 41, 26, 147342) & 302776 \
&(2, 41, 27, 149343) & 306884
end{align*}
Cylindrical algebraic decomposition, using the variable ordering $b,c,a,d$ of your equation gives three solution families:
$a = b = c = 1$ and $d geq 1$, e.g., $(1,1,1,1)$ and $(1,1,1,108)$.
$b = 1$, $c geq 2$, $1 < a < frac{c^3+2}{c^2+2}$, $d = frac{c^3 - a c^2 - 2a + 2}{a-1}$, when $d$ is an integer, e.g., $(2,1,3,7)$ and $(109,1,126,2497)$.
$b geq 2$, $c geq 1$, $1 < a < frac{c^3+2b^3}{c^2+2b^2}$, $d = frac{c^4 - a c^2 + 2 b^3-2ab^2}{a-1}$, when $d$ is an integer, e.g., $(5,2,6,3)$ and $(3,3,100,485,000)$.
(The variable ordering $a,b,c,d$ gives more families, some of which appear to be empty, and most of which require bounds on $b$ and $c$ that are somewhat complicated algebraic functions of the prior variables. The previous version(s) of this answer used that ordering, but the new ordering is used to aid readability.)
$endgroup$
This section of this Answer is responsive to the version of the Question which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d^2) = (2 a^3 + 2 b^3 + c^3 + d^3)$.
Notice that $c$ and $d$ are interchangeable, so we may impose $c leq d$.
One way to look at this is $b,c,d geq 1$ and $a =frac{2 b^3 + c^3 + d^3}{2 b^2 + c^2 + d^2}$, when that is an integer. This leads to the question, does that ever happen? Yes. Notice if $b = c = d$, this is $a = (4b^3)/(4 b^2) = b$, so $a = b = c = d$ is a solution for any choice of $a$.
begin{align*}
&(a,b,c,d) & text{giving} \
&(4,4,4,4) & 384 \
&(5,5,5,5) & 750 \
end{align*}
Cylindrical algebraic decomposition with variable order $c, d, a, b$, gives five cylindrical components. Exhibiting an integer point in a component or showing that a component has no integer points is not easy.
$c geq 1$, $d geq 1$, $a = b = frac{c^3 + d^3}{c^2 + d^2}$, when this is an integer. The fraction is always an integer when $c = d$ and is sometimes an integer otherwise. Examples: $(1,1,1,1)$, $(9,9,5,10)$
$d geq c geq 1$, $a > frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the least real root of $2x^3 -2ax^2 -a(c^2 +d^2) +c^3 + d^3$. Examples: $(9,10,5,5)$, $(17,18,2,14)$, and $(251,255,51,251)$.
$d geq c geq 1$, $a$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when those roots are integers. Having exhausted my tools for determining whether this components has integral points, I am unable to find any or prove that there are none. (Taking intersections with CADs having different variable orders splits this component into pieces, for instance $a geq 2$, $b = 2 a/3$, $c = 2 a/3$, and $d$ is the least real root of $9 x^3 - 9 a x^2 - 4 a^3 = 0$. It's unclear whether that root is ever an integer for an integer $a$ divisible by $3$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(4,2,3,5)$ and $(5,2,2,6)$.
$d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the largest root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(5,4,1,6)$ and $(17,15,5,20)$.
The below is responsive to the first version of the Question, which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d) = (2 a^3 + 2 b^3 + c^3 + d)$.
Well, $a = b = c = d = 1$ is pretty obvious, giving $6$ on both sides.
One infinite family is $a=b=c = 1$, $d$ any positive integer, giving $d+5$. There don't seem to be more with $a = 1$, but I don't know how to prove that.
Here are eight more.
begin{align*}
&(a,b,c,d) & text{giving} \
&(2, 1, 7, 243) & 604 \
&(2, 1, 100, 979998) & 1980016 \
&(2, 2, 8, 384) & 928 \
&(2, 2, 68, 305184) & 619648 \
&(2, 3, 19, 6155) & 13084 \
&(2, 41, 1, 131117) & 268976 \
&(2, 41, 26, 147342) & 302776 \
&(2, 41, 27, 149343) & 306884
end{align*}
Cylindrical algebraic decomposition, using the variable ordering $b,c,a,d$ of your equation gives three solution families:
$a = b = c = 1$ and $d geq 1$, e.g., $(1,1,1,1)$ and $(1,1,1,108)$.
$b = 1$, $c geq 2$, $1 < a < frac{c^3+2}{c^2+2}$, $d = frac{c^3 - a c^2 - 2a + 2}{a-1}$, when $d$ is an integer, e.g., $(2,1,3,7)$ and $(109,1,126,2497)$.
$b geq 2$, $c geq 1$, $1 < a < frac{c^3+2b^3}{c^2+2b^2}$, $d = frac{c^4 - a c^2 + 2 b^3-2ab^2}{a-1}$, when $d$ is an integer, e.g., $(5,2,6,3)$ and $(3,3,100,485,000)$.
(The variable ordering $a,b,c,d$ gives more families, some of which appear to be empty, and most of which require bounds on $b$ and $c$ that are somewhat complicated algebraic functions of the prior variables. The previous version(s) of this answer used that ordering, but the new ordering is used to aid readability.)
edited Jan 15 at 2:56
answered Jan 4 at 21:13
Eric TowersEric Towers
32.8k22370
32.8k22370
$begingroup$
what if $aneq bneq cneq d$?
$endgroup$
– Enzo Creti
Jan 5 at 19:12
$begingroup$
they should be infinite
$endgroup$
– Enzo Creti
Jan 5 at 20:37
$begingroup$
no no ok thank you!
$endgroup$
– Enzo Creti
Jan 5 at 20:40
$begingroup$
@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
$endgroup$
– Eric Towers
Jan 6 at 7:04
add a comment |
$begingroup$
what if $aneq bneq cneq d$?
$endgroup$
– Enzo Creti
Jan 5 at 19:12
$begingroup$
they should be infinite
$endgroup$
– Enzo Creti
Jan 5 at 20:37
$begingroup$
no no ok thank you!
$endgroup$
– Enzo Creti
Jan 5 at 20:40
$begingroup$
@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
$endgroup$
– Eric Towers
Jan 6 at 7:04
$begingroup$
what if $aneq bneq cneq d$?
$endgroup$
– Enzo Creti
Jan 5 at 19:12
$begingroup$
what if $aneq bneq cneq d$?
$endgroup$
– Enzo Creti
Jan 5 at 19:12
$begingroup$
they should be infinite
$endgroup$
– Enzo Creti
Jan 5 at 20:37
$begingroup$
they should be infinite
$endgroup$
– Enzo Creti
Jan 5 at 20:37
$begingroup$
no no ok thank you!
$endgroup$
– Enzo Creti
Jan 5 at 20:40
$begingroup$
no no ok thank you!
$endgroup$
– Enzo Creti
Jan 5 at 20:40
$begingroup$
@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
$endgroup$
– Eric Towers
Jan 6 at 7:04
$begingroup$
@EnzoCreti : Even stronger: $|{a,b,c,d}| = 4$. $(4,2,3,5)$.
$endgroup$
– Eric Towers
Jan 6 at 7:04
add a comment |
$begingroup$
An obvious solution lies where a,b,c,d all equal 1 as:
$$1cdot(2cdot1^2+2cdot1^2+1^2+1) = (2cdot1^3+2cdot1^3+1^3+1)$$
$endgroup$
$begingroup$
this seems a trivial solution
$endgroup$
– user631751
Jan 4 at 20:58
add a comment |
$begingroup$
An obvious solution lies where a,b,c,d all equal 1 as:
$$1cdot(2cdot1^2+2cdot1^2+1^2+1) = (2cdot1^3+2cdot1^3+1^3+1)$$
$endgroup$
$begingroup$
this seems a trivial solution
$endgroup$
– user631751
Jan 4 at 20:58
add a comment |
$begingroup$
An obvious solution lies where a,b,c,d all equal 1 as:
$$1cdot(2cdot1^2+2cdot1^2+1^2+1) = (2cdot1^3+2cdot1^3+1^3+1)$$
$endgroup$
An obvious solution lies where a,b,c,d all equal 1 as:
$$1cdot(2cdot1^2+2cdot1^2+1^2+1) = (2cdot1^3+2cdot1^3+1^3+1)$$
answered Jan 4 at 20:56
Peter ForemanPeter Foreman
2,8771214
2,8771214
$begingroup$
this seems a trivial solution
$endgroup$
– user631751
Jan 4 at 20:58
add a comment |
$begingroup$
this seems a trivial solution
$endgroup$
– user631751
Jan 4 at 20:58
$begingroup$
this seems a trivial solution
$endgroup$
– user631751
Jan 4 at 20:58
$begingroup$
this seems a trivial solution
$endgroup$
– user631751
Jan 4 at 20:58
add a comment |
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$begingroup$
Do you mean different values? If not, take $a=b=c=d=1$.
$endgroup$
– AugSB
Jan 4 at 20:52
$begingroup$
As a rough estimate based on the prime number theorem there ought to be about three $k$-digit numbers such that the base-ten concatenation of $2^{n+1}-1$ and $2^n-1$ is prime, for each $k$.
$endgroup$
– Henning Makholm
Jan 4 at 20:58
1
$begingroup$
A suggestion: replace the two $d$'s in the equation you ask about with a $d^2$ on the left and a $d^3$ on the right.
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– Barry Cipra
Jan 4 at 21:08
$begingroup$
Why not make this homogeneous on each side since $1=1^2=1^3$? This is equivalent to @BarryCipra's comment.
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– Mark Bennet
Jan 4 at 21:13
$begingroup$
@Barry Cipra now i replaced as you suggested
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– user631751
Jan 4 at 21:33