Proof: $n^2 - 7$ is not divisble by 5












0














I tried to prove that $n^2 - 7$ is not divisible by $5$ via proof by contradiction. Does this look right?



Suppose $n^2 - 7$ is divisible by $5$. Then:



$n^2 - 7 = 5g$, $g in mathbb{Z}$.
$n^2 = 5g + 7$.



Consider the case where $n$ is even.



$(2x)^2 = 5g + 7$, $x in mathbb{Z}$.



$4x^2 = 5g + 7$.



$4s = 5g + 7$, $s = x^2, s in mathbb{Z}$ as integers are closed under multiplication.



$2s$ is even, and $5g + 7$ is odd if we consider that g is an even number. so we have a contradiction.



Consider the case where $n$ is odd.



$(2x + 1)^2 = 5g + 7$, $x in mathbb{Z}$



$4x^2 + 4x = 5g + 6$



$4x^2 + 4x = 5g + 6$



$4(x^2 + x) = 5g + 6$



$4j = 5g + 6$, $j = x^2 + x, j in mathbb{Z}$ as integers are closed under addition



$2d = 5g + 7$, $d = 2j; d in mathbb{Z}$ as integers are closed under multiplication



$2d$ is even, and $5g + 7$ is odd if we consider that g is an odd number. An even number cannot equal an odd number, so we have a contradiction.



As both cases have a contradiction, the original supposition is false, and $n^2 - 7$ is not divisible by $5$.



is my proof correct because I cannot prove that $5g + 6$ or $5g + 7$ are odd so I assumed that in the even case g is even and in the odd case g is odd










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  • You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4?
    – user247327
    Nov 24 at 13:00










  • sorry i meant Suppose $n^2−7$ is divisible by $5$
    – Omaroo Baniessa
    Nov 24 at 13:37


















0














I tried to prove that $n^2 - 7$ is not divisible by $5$ via proof by contradiction. Does this look right?



Suppose $n^2 - 7$ is divisible by $5$. Then:



$n^2 - 7 = 5g$, $g in mathbb{Z}$.
$n^2 = 5g + 7$.



Consider the case where $n$ is even.



$(2x)^2 = 5g + 7$, $x in mathbb{Z}$.



$4x^2 = 5g + 7$.



$4s = 5g + 7$, $s = x^2, s in mathbb{Z}$ as integers are closed under multiplication.



$2s$ is even, and $5g + 7$ is odd if we consider that g is an even number. so we have a contradiction.



Consider the case where $n$ is odd.



$(2x + 1)^2 = 5g + 7$, $x in mathbb{Z}$



$4x^2 + 4x = 5g + 6$



$4x^2 + 4x = 5g + 6$



$4(x^2 + x) = 5g + 6$



$4j = 5g + 6$, $j = x^2 + x, j in mathbb{Z}$ as integers are closed under addition



$2d = 5g + 7$, $d = 2j; d in mathbb{Z}$ as integers are closed under multiplication



$2d$ is even, and $5g + 7$ is odd if we consider that g is an odd number. An even number cannot equal an odd number, so we have a contradiction.



As both cases have a contradiction, the original supposition is false, and $n^2 - 7$ is not divisible by $5$.



is my proof correct because I cannot prove that $5g + 6$ or $5g + 7$ are odd so I assumed that in the even case g is even and in the odd case g is odd










share|cite|improve this question
























  • You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4?
    – user247327
    Nov 24 at 13:00










  • sorry i meant Suppose $n^2−7$ is divisible by $5$
    – Omaroo Baniessa
    Nov 24 at 13:37
















0












0








0







I tried to prove that $n^2 - 7$ is not divisible by $5$ via proof by contradiction. Does this look right?



Suppose $n^2 - 7$ is divisible by $5$. Then:



$n^2 - 7 = 5g$, $g in mathbb{Z}$.
$n^2 = 5g + 7$.



Consider the case where $n$ is even.



$(2x)^2 = 5g + 7$, $x in mathbb{Z}$.



$4x^2 = 5g + 7$.



$4s = 5g + 7$, $s = x^2, s in mathbb{Z}$ as integers are closed under multiplication.



$2s$ is even, and $5g + 7$ is odd if we consider that g is an even number. so we have a contradiction.



Consider the case where $n$ is odd.



$(2x + 1)^2 = 5g + 7$, $x in mathbb{Z}$



$4x^2 + 4x = 5g + 6$



$4x^2 + 4x = 5g + 6$



$4(x^2 + x) = 5g + 6$



$4j = 5g + 6$, $j = x^2 + x, j in mathbb{Z}$ as integers are closed under addition



$2d = 5g + 7$, $d = 2j; d in mathbb{Z}$ as integers are closed under multiplication



$2d$ is even, and $5g + 7$ is odd if we consider that g is an odd number. An even number cannot equal an odd number, so we have a contradiction.



As both cases have a contradiction, the original supposition is false, and $n^2 - 7$ is not divisible by $5$.



is my proof correct because I cannot prove that $5g + 6$ or $5g + 7$ are odd so I assumed that in the even case g is even and in the odd case g is odd










share|cite|improve this question















I tried to prove that $n^2 - 7$ is not divisible by $5$ via proof by contradiction. Does this look right?



Suppose $n^2 - 7$ is divisible by $5$. Then:



$n^2 - 7 = 5g$, $g in mathbb{Z}$.
$n^2 = 5g + 7$.



Consider the case where $n$ is even.



$(2x)^2 = 5g + 7$, $x in mathbb{Z}$.



$4x^2 = 5g + 7$.



$4s = 5g + 7$, $s = x^2, s in mathbb{Z}$ as integers are closed under multiplication.



$2s$ is even, and $5g + 7$ is odd if we consider that g is an even number. so we have a contradiction.



Consider the case where $n$ is odd.



$(2x + 1)^2 = 5g + 7$, $x in mathbb{Z}$



$4x^2 + 4x = 5g + 6$



$4x^2 + 4x = 5g + 6$



$4(x^2 + x) = 5g + 6$



$4j = 5g + 6$, $j = x^2 + x, j in mathbb{Z}$ as integers are closed under addition



$2d = 5g + 7$, $d = 2j; d in mathbb{Z}$ as integers are closed under multiplication



$2d$ is even, and $5g + 7$ is odd if we consider that g is an odd number. An even number cannot equal an odd number, so we have a contradiction.



As both cases have a contradiction, the original supposition is false, and $n^2 - 7$ is not divisible by $5$.



is my proof correct because I cannot prove that $5g + 6$ or $5g + 7$ are odd so I assumed that in the even case g is even and in the odd case g is odd







proof-verification






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edited Nov 24 at 13:36

























asked Nov 24 at 12:55









Omaroo Baniessa

61




61












  • You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4?
    – user247327
    Nov 24 at 13:00










  • sorry i meant Suppose $n^2−7$ is divisible by $5$
    – Omaroo Baniessa
    Nov 24 at 13:37




















  • You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4?
    – user247327
    Nov 24 at 13:00










  • sorry i meant Suppose $n^2−7$ is divisible by $5$
    – Omaroo Baniessa
    Nov 24 at 13:37


















You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4?
– user247327
Nov 24 at 13:00




You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4?
– user247327
Nov 24 at 13:00












sorry i meant Suppose $n^2−7$ is divisible by $5$
– Omaroo Baniessa
Nov 24 at 13:37






sorry i meant Suppose $n^2−7$ is divisible by $5$
– Omaroo Baniessa
Nov 24 at 13:37












6 Answers
6






active

oldest

votes


















5














It seems rather overcomplicated. Instead, simply note that the question is equivalent to solving $n^2 - 2 = 0$ in $mathbb{Z}/5mathbb{Z}$. But there are only five values in $mathbb{Z}/5mathbb{Z}$, and we can just try them all to see if they're solutions:



begin{align*}
0^2 - 2 &= -2 neq 0\
1^2 - 2 &= -1 neq 0\
2^2 - 2 &= 2 neq 0\
(-1)^2-2 &= -1 neq 0\
(-2)^2-2 &= 2 neq 0.
end{align*}



None of them are solutions, so there are no solutions to the original problem.






share|cite|improve this answer





















  • sorry i meant Suppose n2−7 is divisible by 5
    – Omaroo Baniessa
    Nov 24 at 18:35










  • I know, and the above precisely proves that there is no $n$ with that property.
    – user3482749
    Nov 24 at 18:37



















2














Suppose $n^2-7$ is divisible by $5$ i.e. $n^2equiv 7 mod 5implies n^2equiv 2mod 5$. We can check $:forall ninmathbb{Z}quad n^2equivbegin{cases}0 &mod5\1 &mod5\4 &mod5end{cases} $. So $n^2-7$ is not divisible by $5$.






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    1














    Here is an alternate approach. $n^2-7$ is divisible by 5 only if the last digit of $n^2-7$ is either 0 or 5. This means the last digit of $n^2$ is either 7 or 2. Now can you complete the proof?






    share|cite|improve this answer





























      1














      Your proof is incorrect. Note $,n^2 = 5g+7 ,Rightarrowbmod 2!: nequiv g+1,$ so $n$ and $g$ have opposite parity. But your proof only considers the cases when they have equal parity.



      Instead note $bmod 5!: nequiv 0,pm1,pm2,Rightarrow,n^2equiv 0,1,4 notequiv 7$



      Alternatively $,n^2equiv 7equiv 2,Rightarrow, n^4equiv 2^2notequiv 1,$ contra little Fermat.






      share|cite|improve this answer





























        0














        Want to show $n^2-7 neq 0 (mod 5)$ which is the same as show $n^2-2 neq 0 (mod5)$. Here $0^2=0 (mod5), 1^2=1 (mod5), 2^2=4 (mod5), 3^2=4 (mod5), 4^2=1 (mod5)$.
        Then, that means $n^2=1, 4 (mod5)$. SO if it is $1$, then $n^2-2=-1(mod5)$ and if $n^2=4$, $n^2-1=3(mod5)$. In all cases, $n^2-2neq 0(mod5)$.






        share|cite|improve this answer





























          0














          If possible let $ n^2-7$ is divisible by $5$, then we have



          $n^2-7=5k, k in mathbb{Z}, ........(1)$.



          Now let $ k in mathbb{Z}^{+}$, then for $N ( neq 7) in mathbb{N}$, we have $k=n=N$ ( without loss of generality)



          Then from $(1)$, we get



          $ N^2-7=5N \ Rightarrow N^2-5N-7=0 \ Rightarrow N=7, -2$



          But as $N neq 7$, we must have $ N=-2 in mathbb{N}$=set of natural number, which is impossible.



          Thus $n^2-7$ is not divisible by $5$.






          share|cite|improve this answer





















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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            It seems rather overcomplicated. Instead, simply note that the question is equivalent to solving $n^2 - 2 = 0$ in $mathbb{Z}/5mathbb{Z}$. But there are only five values in $mathbb{Z}/5mathbb{Z}$, and we can just try them all to see if they're solutions:



            begin{align*}
            0^2 - 2 &= -2 neq 0\
            1^2 - 2 &= -1 neq 0\
            2^2 - 2 &= 2 neq 0\
            (-1)^2-2 &= -1 neq 0\
            (-2)^2-2 &= 2 neq 0.
            end{align*}



            None of them are solutions, so there are no solutions to the original problem.






            share|cite|improve this answer





















            • sorry i meant Suppose n2−7 is divisible by 5
              – Omaroo Baniessa
              Nov 24 at 18:35










            • I know, and the above precisely proves that there is no $n$ with that property.
              – user3482749
              Nov 24 at 18:37
















            5














            It seems rather overcomplicated. Instead, simply note that the question is equivalent to solving $n^2 - 2 = 0$ in $mathbb{Z}/5mathbb{Z}$. But there are only five values in $mathbb{Z}/5mathbb{Z}$, and we can just try them all to see if they're solutions:



            begin{align*}
            0^2 - 2 &= -2 neq 0\
            1^2 - 2 &= -1 neq 0\
            2^2 - 2 &= 2 neq 0\
            (-1)^2-2 &= -1 neq 0\
            (-2)^2-2 &= 2 neq 0.
            end{align*}



            None of them are solutions, so there are no solutions to the original problem.






            share|cite|improve this answer





















            • sorry i meant Suppose n2−7 is divisible by 5
              – Omaroo Baniessa
              Nov 24 at 18:35










            • I know, and the above precisely proves that there is no $n$ with that property.
              – user3482749
              Nov 24 at 18:37














            5












            5








            5






            It seems rather overcomplicated. Instead, simply note that the question is equivalent to solving $n^2 - 2 = 0$ in $mathbb{Z}/5mathbb{Z}$. But there are only five values in $mathbb{Z}/5mathbb{Z}$, and we can just try them all to see if they're solutions:



            begin{align*}
            0^2 - 2 &= -2 neq 0\
            1^2 - 2 &= -1 neq 0\
            2^2 - 2 &= 2 neq 0\
            (-1)^2-2 &= -1 neq 0\
            (-2)^2-2 &= 2 neq 0.
            end{align*}



            None of them are solutions, so there are no solutions to the original problem.






            share|cite|improve this answer












            It seems rather overcomplicated. Instead, simply note that the question is equivalent to solving $n^2 - 2 = 0$ in $mathbb{Z}/5mathbb{Z}$. But there are only five values in $mathbb{Z}/5mathbb{Z}$, and we can just try them all to see if they're solutions:



            begin{align*}
            0^2 - 2 &= -2 neq 0\
            1^2 - 2 &= -1 neq 0\
            2^2 - 2 &= 2 neq 0\
            (-1)^2-2 &= -1 neq 0\
            (-2)^2-2 &= 2 neq 0.
            end{align*}



            None of them are solutions, so there are no solutions to the original problem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 13:03









            user3482749

            2,226414




            2,226414












            • sorry i meant Suppose n2−7 is divisible by 5
              – Omaroo Baniessa
              Nov 24 at 18:35










            • I know, and the above precisely proves that there is no $n$ with that property.
              – user3482749
              Nov 24 at 18:37


















            • sorry i meant Suppose n2−7 is divisible by 5
              – Omaroo Baniessa
              Nov 24 at 18:35










            • I know, and the above precisely proves that there is no $n$ with that property.
              – user3482749
              Nov 24 at 18:37
















            sorry i meant Suppose n2−7 is divisible by 5
            – Omaroo Baniessa
            Nov 24 at 18:35




            sorry i meant Suppose n2−7 is divisible by 5
            – Omaroo Baniessa
            Nov 24 at 18:35












            I know, and the above precisely proves that there is no $n$ with that property.
            – user3482749
            Nov 24 at 18:37




            I know, and the above precisely proves that there is no $n$ with that property.
            – user3482749
            Nov 24 at 18:37











            2














            Suppose $n^2-7$ is divisible by $5$ i.e. $n^2equiv 7 mod 5implies n^2equiv 2mod 5$. We can check $:forall ninmathbb{Z}quad n^2equivbegin{cases}0 &mod5\1 &mod5\4 &mod5end{cases} $. So $n^2-7$ is not divisible by $5$.






            share|cite|improve this answer


























              2














              Suppose $n^2-7$ is divisible by $5$ i.e. $n^2equiv 7 mod 5implies n^2equiv 2mod 5$. We can check $:forall ninmathbb{Z}quad n^2equivbegin{cases}0 &mod5\1 &mod5\4 &mod5end{cases} $. So $n^2-7$ is not divisible by $5$.






              share|cite|improve this answer
























                2












                2








                2






                Suppose $n^2-7$ is divisible by $5$ i.e. $n^2equiv 7 mod 5implies n^2equiv 2mod 5$. We can check $:forall ninmathbb{Z}quad n^2equivbegin{cases}0 &mod5\1 &mod5\4 &mod5end{cases} $. So $n^2-7$ is not divisible by $5$.






                share|cite|improve this answer












                Suppose $n^2-7$ is divisible by $5$ i.e. $n^2equiv 7 mod 5implies n^2equiv 2mod 5$. We can check $:forall ninmathbb{Z}quad n^2equivbegin{cases}0 &mod5\1 &mod5\4 &mod5end{cases} $. So $n^2-7$ is not divisible by $5$.







                share|cite|improve this answer












                share|cite|improve this answer



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                answered Nov 24 at 13:07









                Yadati Kiran

                1,688519




                1,688519























                    1














                    Here is an alternate approach. $n^2-7$ is divisible by 5 only if the last digit of $n^2-7$ is either 0 or 5. This means the last digit of $n^2$ is either 7 or 2. Now can you complete the proof?






                    share|cite|improve this answer


























                      1














                      Here is an alternate approach. $n^2-7$ is divisible by 5 only if the last digit of $n^2-7$ is either 0 or 5. This means the last digit of $n^2$ is either 7 or 2. Now can you complete the proof?






                      share|cite|improve this answer
























                        1












                        1








                        1






                        Here is an alternate approach. $n^2-7$ is divisible by 5 only if the last digit of $n^2-7$ is either 0 or 5. This means the last digit of $n^2$ is either 7 or 2. Now can you complete the proof?






                        share|cite|improve this answer












                        Here is an alternate approach. $n^2-7$ is divisible by 5 only if the last digit of $n^2-7$ is either 0 or 5. This means the last digit of $n^2$ is either 7 or 2. Now can you complete the proof?







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 24 at 13:06









                        Thomas Shelby

                        1,185116




                        1,185116























                            1














                            Your proof is incorrect. Note $,n^2 = 5g+7 ,Rightarrowbmod 2!: nequiv g+1,$ so $n$ and $g$ have opposite parity. But your proof only considers the cases when they have equal parity.



                            Instead note $bmod 5!: nequiv 0,pm1,pm2,Rightarrow,n^2equiv 0,1,4 notequiv 7$



                            Alternatively $,n^2equiv 7equiv 2,Rightarrow, n^4equiv 2^2notequiv 1,$ contra little Fermat.






                            share|cite|improve this answer


























                              1














                              Your proof is incorrect. Note $,n^2 = 5g+7 ,Rightarrowbmod 2!: nequiv g+1,$ so $n$ and $g$ have opposite parity. But your proof only considers the cases when they have equal parity.



                              Instead note $bmod 5!: nequiv 0,pm1,pm2,Rightarrow,n^2equiv 0,1,4 notequiv 7$



                              Alternatively $,n^2equiv 7equiv 2,Rightarrow, n^4equiv 2^2notequiv 1,$ contra little Fermat.






                              share|cite|improve this answer
























                                1












                                1








                                1






                                Your proof is incorrect. Note $,n^2 = 5g+7 ,Rightarrowbmod 2!: nequiv g+1,$ so $n$ and $g$ have opposite parity. But your proof only considers the cases when they have equal parity.



                                Instead note $bmod 5!: nequiv 0,pm1,pm2,Rightarrow,n^2equiv 0,1,4 notequiv 7$



                                Alternatively $,n^2equiv 7equiv 2,Rightarrow, n^4equiv 2^2notequiv 1,$ contra little Fermat.






                                share|cite|improve this answer












                                Your proof is incorrect. Note $,n^2 = 5g+7 ,Rightarrowbmod 2!: nequiv g+1,$ so $n$ and $g$ have opposite parity. But your proof only considers the cases when they have equal parity.



                                Instead note $bmod 5!: nequiv 0,pm1,pm2,Rightarrow,n^2equiv 0,1,4 notequiv 7$



                                Alternatively $,n^2equiv 7equiv 2,Rightarrow, n^4equiv 2^2notequiv 1,$ contra little Fermat.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 24 at 14:28









                                Bill Dubuque

                                208k29190626




                                208k29190626























                                    0














                                    Want to show $n^2-7 neq 0 (mod 5)$ which is the same as show $n^2-2 neq 0 (mod5)$. Here $0^2=0 (mod5), 1^2=1 (mod5), 2^2=4 (mod5), 3^2=4 (mod5), 4^2=1 (mod5)$.
                                    Then, that means $n^2=1, 4 (mod5)$. SO if it is $1$, then $n^2-2=-1(mod5)$ and if $n^2=4$, $n^2-1=3(mod5)$. In all cases, $n^2-2neq 0(mod5)$.






                                    share|cite|improve this answer


























                                      0














                                      Want to show $n^2-7 neq 0 (mod 5)$ which is the same as show $n^2-2 neq 0 (mod5)$. Here $0^2=0 (mod5), 1^2=1 (mod5), 2^2=4 (mod5), 3^2=4 (mod5), 4^2=1 (mod5)$.
                                      Then, that means $n^2=1, 4 (mod5)$. SO if it is $1$, then $n^2-2=-1(mod5)$ and if $n^2=4$, $n^2-1=3(mod5)$. In all cases, $n^2-2neq 0(mod5)$.






                                      share|cite|improve this answer
























                                        0












                                        0








                                        0






                                        Want to show $n^2-7 neq 0 (mod 5)$ which is the same as show $n^2-2 neq 0 (mod5)$. Here $0^2=0 (mod5), 1^2=1 (mod5), 2^2=4 (mod5), 3^2=4 (mod5), 4^2=1 (mod5)$.
                                        Then, that means $n^2=1, 4 (mod5)$. SO if it is $1$, then $n^2-2=-1(mod5)$ and if $n^2=4$, $n^2-1=3(mod5)$. In all cases, $n^2-2neq 0(mod5)$.






                                        share|cite|improve this answer












                                        Want to show $n^2-7 neq 0 (mod 5)$ which is the same as show $n^2-2 neq 0 (mod5)$. Here $0^2=0 (mod5), 1^2=1 (mod5), 2^2=4 (mod5), 3^2=4 (mod5), 4^2=1 (mod5)$.
                                        Then, that means $n^2=1, 4 (mod5)$. SO if it is $1$, then $n^2-2=-1(mod5)$ and if $n^2=4$, $n^2-1=3(mod5)$. In all cases, $n^2-2neq 0(mod5)$.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Nov 24 at 13:07









                                        mathnoob

                                        1,775322




                                        1,775322























                                            0














                                            If possible let $ n^2-7$ is divisible by $5$, then we have



                                            $n^2-7=5k, k in mathbb{Z}, ........(1)$.



                                            Now let $ k in mathbb{Z}^{+}$, then for $N ( neq 7) in mathbb{N}$, we have $k=n=N$ ( without loss of generality)



                                            Then from $(1)$, we get



                                            $ N^2-7=5N \ Rightarrow N^2-5N-7=0 \ Rightarrow N=7, -2$



                                            But as $N neq 7$, we must have $ N=-2 in mathbb{N}$=set of natural number, which is impossible.



                                            Thus $n^2-7$ is not divisible by $5$.






                                            share|cite|improve this answer


























                                              0














                                              If possible let $ n^2-7$ is divisible by $5$, then we have



                                              $n^2-7=5k, k in mathbb{Z}, ........(1)$.



                                              Now let $ k in mathbb{Z}^{+}$, then for $N ( neq 7) in mathbb{N}$, we have $k=n=N$ ( without loss of generality)



                                              Then from $(1)$, we get



                                              $ N^2-7=5N \ Rightarrow N^2-5N-7=0 \ Rightarrow N=7, -2$



                                              But as $N neq 7$, we must have $ N=-2 in mathbb{N}$=set of natural number, which is impossible.



                                              Thus $n^2-7$ is not divisible by $5$.






                                              share|cite|improve this answer
























                                                0












                                                0








                                                0






                                                If possible let $ n^2-7$ is divisible by $5$, then we have



                                                $n^2-7=5k, k in mathbb{Z}, ........(1)$.



                                                Now let $ k in mathbb{Z}^{+}$, then for $N ( neq 7) in mathbb{N}$, we have $k=n=N$ ( without loss of generality)



                                                Then from $(1)$, we get



                                                $ N^2-7=5N \ Rightarrow N^2-5N-7=0 \ Rightarrow N=7, -2$



                                                But as $N neq 7$, we must have $ N=-2 in mathbb{N}$=set of natural number, which is impossible.



                                                Thus $n^2-7$ is not divisible by $5$.






                                                share|cite|improve this answer












                                                If possible let $ n^2-7$ is divisible by $5$, then we have



                                                $n^2-7=5k, k in mathbb{Z}, ........(1)$.



                                                Now let $ k in mathbb{Z}^{+}$, then for $N ( neq 7) in mathbb{N}$, we have $k=n=N$ ( without loss of generality)



                                                Then from $(1)$, we get



                                                $ N^2-7=5N \ Rightarrow N^2-5N-7=0 \ Rightarrow N=7, -2$



                                                But as $N neq 7$, we must have $ N=-2 in mathbb{N}$=set of natural number, which is impossible.



                                                Thus $n^2-7$ is not divisible by $5$.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Nov 24 at 14:17









                                                M. A. SARKAR

                                                2,1271619




                                                2,1271619






























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