I have solved the problem . I am posting it handwritten shortly .

Pls verify it .

That solves only the case $n$ even, because if $a$ is not a perfect square, then $a^{n/2} \notin \mathbb{Z}$ when $n$ is odd.

Here is the argument for odd $n$: Let $n$ be odd, $a^n-2^n=k^2$, and suppose that both $a$ is odd, as well. Since $a^n-2^n$ is odd, it follows that $a^n\equiv 1\pmod{4}$. Consequently, $a\equiv 1\pmod{4}$. Now, observe that, $a^n+2^n = k^2+2^{n+1}$, and since $n$ is odd, it holds that $a+2\mid k^2+2^{n+1}$. Now, since $a+2\equiv 3\pmod{4}$, there exists a prime number $q\equiv 3\pmod{4}$, such that $q\mid a+2$, consequently $q\mid k^2+\left(2^{\frac{n+1}{2}}\right)^2$. But since $-1$ is not a quadratic residue modulo such prime, and $q$ clearly cannot divide $2^{(n+1)/2}$, this yields a contradiction.

grupyorum wrote: Here is the argument for odd $n$: Let $n$ be odd, $a^n-2^n=k^2$, and suppose that both $a$ is odd, as well. Since $a^n-2^n$ is odd, it follows that $a^n\equiv 1\pmod{4}$. Consequently, $a\equiv 1\pmod{4}$. Now, observe that, $a^n+2^n = k^2+2^{n+1}$, and since $n$ is odd, it holds that $a+2\mid k^2+2^{n+1}$. Now, since $a+2\equiv 3\pmod{4}$, there exists a prime number $q\equiv 3\pmod{4}$, such that $q\mid a+2$, consequently $q\mid k^2+\left(2^{\frac{n+1}{2}}\right)^2$. But since $-1$ is not a quadratic residue modulo such prime, and $q$ clearly cannot divide $2^{(n+1)/2}$, this yields a contradiction. Could you please describe your last statement about the quardratic residue modulo \(q\)? why it has to be only \(-1\) and not \(-3,..\)

Junio wrote: grupyorum wrote: Here is the argument for odd $n$: Let $n$ be odd, $a^n-2^n=k^2$, and suppose that both $a$ is odd, as well. Since $a^n-2^n$ is odd, it follows that $a^n\equiv 1\pmod{4}$. Consequently, $a\equiv 1\pmod{4}$. Now, observe that, $a^n+2^n = k^2+2^{n+1}$, and since $n$ is odd, it holds that $a+2\mid k^2+2^{n+1}$. Now, since $a+2\equiv 3\pmod{4}$, there exists a prime number $q\equiv 3\pmod{4}$, such that $q\mid a+2$, consequently $q\mid k^2+\left(2^{\frac{n+1}{2}}\right)^2$. But since $-1$ is not a quadratic residue modulo such prime, and $q$ clearly cannot divide $2^{(n+1)/2}$, this yields a contradiction. Could you please describe your last statement about the quardratic residue modulo \(q\)? why it has to be only \(-1\) and not \(-3,..\) It is said that there EXISTS a $q \equiv 3(mod 4)$. So if we take that prime it will divide the sum of two squares. Rest is stated above.

electrovector wrote: Junio wrote: grupyorum wrote: Here is the argument for odd $n$: Let $n$ be odd, $a^n-2^n=k^2$, and suppose that both $a$ is odd, as well. Since $a^n-2^n$ is odd, it follows that $a^n\equiv 1\pmod{4}$. Consequently, $a\equiv 1\pmod{4}$. Now, observe that, $a^n+2^n = k^2+2^{n+1}$, and since $n$ is odd, it holds that $a+2\mid k^2+2^{n+1}$. Now, since $a+2\equiv 3\pmod{4}$, there exists a prime number $q\equiv 3\pmod{4}$, such that $q\mid a+2$, consequently $q\mid k^2+\left(2^{\frac{n+1}{2}}\right)^2$. But since $-1$ is not a quadratic residue modulo such prime, and $q$ clearly cannot divide $2^{(n+1)/2}$, this yields a contradiction. Could you please describe your last statement about the quardratic residue modulo \(q\)? why it has to be only \(-1\) and not \(-3,..\) It is said that there EXISTS a $q \equiv 3(mod 4)$. So if we take that prime it will divide the sum of two squares. Rest is stated above. I understood that \(q|\left(k^2+\left(2^{\frac{n+1}{2}}\right)^2\right)\), but what I didn't is that, how come Mr.grupyorum conclude, in this case for the right hand side of the equivalence, \(\left(2^{\frac{n+1}{2}}\right)^2\equiv 1 \pmod q\) which is \(k^2\equiv -1 \pmod q\) (and it was been proven that there exists not such a case here), to be \(-1\) and not to be \(-3,-5,...\) the other cases.

theorem