Prove that $2^n+3^n $ is never a perfect square

My attempt :

If $n$ is odd, then the square must be 2 (mod 3), which is not possible.

Hence $n =2m$

$2^{2m}+3^{2m}=(2^m+a)^2$

$a^2+2^{m+1}a=3^{2m}$

$a (a+2^{m+1})=3^{2m} $

By fundamental theorem of arithmetic,

$a=3^x $

$3^x +2^{m+1}=3^y $

$2^{m+1}=3^x (3^{y-x}-1) $

Which is not possible by Fundamental theorem of Arithmetic

Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.

Solution 1:

Work $\pmod 3$. We see that $$2^n+3^n\equiv 2^n\pmod 3$$ from which we quickly deduce that your expression is only a square $\pmod 3$ if $n$ is even.

Now work $\pmod 5$. We see that $$2^n+3^n\equiv 2^n+(-2)^n\equiv 2^n(1+(-1)^n)\pmod 5$$ from which we deduce that your expression is only a square $\pmod 5$ if $n$ is odd.

Solution 2:

Your method is slightly wrong, as you have to deal with the $x=0$ case. However, the remaining diophantine equation is simple.

Here, I provide you with an alternative approach, though admittedly yours seems better.

For the case $m \equiv 0 \pmod 2$, note that $$2^{2m}+3^{2m}=n^2$$ implies that $2^m=2ab$, and $3^m=a^2-b^2$ where $\gcd(a,b)=1, a \not \equiv b \pmod 2$. This implies that $a=2^{m-1}, b=1$. So all that remains is finding $m,y$ such that $3^{m}=2^{2m-2}-1$.

Solution 3:

This is more convoluted than your way, but I already wrote it so I'll just leave it here :-)

$4^m = (a-3^m)(a+3^m) \implies a-3^m = 2^p, a+3^m = 2^q$ with $p+q = 2m$. Note that $q \ge p$

Then $a = 2^p + 3^m = 2^q - 3^m \iff 2^p (2^{q-p}-1) = 2\cdot 3^m$

But then it must be $p=1$, $q = 2m-1$, and we have

$$2^{2(m-1)}-1 = (2^{m-1}-1)(2^{m-1}+1) = 3^m$$

Since only one of the factors in the LHS can be divisible by $3$, it means the other one must be $1$, hence we get $2^{m-1}-1 = 1 \implies m = 2$

But in this case $2^{m-1}+1 = 3 \neq 3^2 $, so this is impossible.

Solution 4:

Yet another approach.

For the odd case of $n$, as you mentioned the result is $2\mod 3$ which cannot be a square number.

As for the even case $n=2m$, we have $2^{2m}+3^{2m}\mod 10$ being equal to $2$ (for when $m=0$) or either $3$ (for $n=2+4k,k\geq0$) or $7$ (for $n=4k,k\geq1$), which clearly cannot be square (as a square number modulus $10$ takes values $0,1,4,9,6,5$).