Prove that there exist no natural number k such that $3^k+5^k$ is a square of an integer number? How to prove that there exist no natural number $k$ such that $3^k+5^k$ is a square of an integer number
 A: If $k$ is odd we get that $3^k+5^k \equiv 2 \pmod{3}$, which isn't possible in a square.
As the result will be even, $4n^2= 3^k+5^k$
If $k$ is even, $3^k+5^k \equiv 1^k+(-1)^k \equiv 2 \pmod{4}$. But $4n^2 = 3^k+5^k \Longrightarrow 4 \mid 3^k+5^k$. So we obtain a contradiction. 
A: Tip: You can see that $3^k + 5^k$ is never odd but even for any $k$. So you have to prove that there are no solutons to
$$ (2n)^2 = 3^k + 5^k $$
where $n$ is a positive integer. Do you think you can give it a try now?
A: Even values of $k \rightarrow \equiv 2 \bmod 8$.
Odd values of $k \rightarrow \equiv 2 \bmod 3$.
A: An elementary solution without modular arithmetic. :)
If $k$ is odd, then write
$$a^2 = 3^k+5^k= 8(3^{k-1}-3^{k-2}5+...+5^{k-1})$$
Since expression in bracket is odd (we have odd odd numbers) we have $8|a^2$ but $16\not|a^2$. A contradiction.
Say $k$ is even. Then $k=2n$ so $(a-3^n)(a+3^n)= 5^{2n}$, so $a-3^n= 5^{x}$ and $a+3^n= 5^{y}$ for some non negative integer $x$ and $y$ (we see that $y>x$ and $x+y=2n$) thus:
$$ 2\cdot 3^n = 5^y-5^x= 5^x(5^{y-x}-1)$$
so $x=0$. So $2\cdot 3^n = 5^{2n}-1$ which is impossible since that left side is smaller then right side for each $n$ (easy to prove with induction).
A: We can see easily, that $3^k+5^k$ is even for every $k$. When you square an odd number the result is odd. So the number you search has to be even too and you get 
$3^k+5^k=(2n)^2=4n^2$. Hence 4 divides $3^k+5^k$. Now we observe 
$3^k+5^k\mod 4\equiv (-1)^k+(1)^k\equiv\begin{cases}0,~~ k=2m+1\\ 2,~~ k=2m\end{cases} $
This can not be.
A: Hint : You can consider using congruences
Modulo 3 and modulo 4
Or
Modulo 8
