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Find all positive integers m, n such that $6^m + 2^n + 2$ is a perfect square. I've tried keeping a constant value of m and finding out n. Eg:

$m=1, n=0$

$m=1, n=3$

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    $\begingroup$ Here is a hint: Unless $m= 0$ or $n= 0$ then $6^m + 2^n + 2$ is even so if it is a square it is a square of an even number. $\endgroup$ – Alex J Best Apr 19 '19 at 8:40
  • $\begingroup$ Compare with this question, or this one, or this one. $\endgroup$ – Dietrich Burde Apr 19 '19 at 8:56
  • $\begingroup$ Another hint : If $m>1$ and $n>1$ , the number is of the form $4k+2$ , hence not a perfect square. Usually $0$ is not considered positive. $\endgroup$ – Peter Apr 19 '19 at 13:01
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As people in the comments mentioned, if $m > 1$ and $n > 1$ than our number divide $2$, but not divide $4$, so it cant be square. So the only cases are:

$1$) $m = 1$ In this case we have equation $2^n + 8$ is a perfect square. But if $n > 3$ than this number divide $8$, but not divide $16$, so it cant be square. So just consider $n = {1,2,3}$

$2$) $n = 1$ In this case we have equation $6^m + 4$ is a perfect square. But $6^m + 4 \equiv$ $3\ or \ 5\ (mod\ 7)$ which is not quadratic residue.

Thats all cases.

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  • $\begingroup$ Numbers are positive integers or non-negative? $\endgroup$ – Arial Pilisov Apr 19 '19 at 18:43
  • $\begingroup$ If you need non-negative integers than: 1) $m = 0\\$. $2^n+3 \equiv 3 \ (mod\ 4)$ which in not quadratic residue. 2)$n = 0\\$ $6^m+3$ is square. If $m>1$ than this equation divide $3$, but not divide $9$, so it cant be square. Than just consider $m=0\ or\ 1$ $\endgroup$ – Arial Pilisov Apr 19 '19 at 18:48
  • $\begingroup$ I needed the postive integers only. Thank you so much. $\endgroup$ – Tapi Apr 20 '19 at 15:49
  • $\begingroup$ In Case 1, shouldn't we show that when a number is divisible by a prime but not by the prime square, then it is not a perfect square? However, you've shown that it is divisible by "8" but not by 16. Shouldn't it be 4 instead of 8 as the square of 4 is 16? Also, 4 is not a prime number. So is the case still justified considering I cannot find any n>3? $\endgroup$ – Tapi Apr 22 '19 at 19:26
  • $\begingroup$ I showed that the number is divisible $2^3$, but not divisible by $2^4$, so it cant be square, because if the number is square than every prime that divide our square has even power. $\endgroup$ – Arial Pilisov Apr 23 '19 at 11:19

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