Prove that there exist no natural number k such that $3^k+5^k$ is a square of an integer number?

Question

How to prove that there exist no natural number $k$ such that $3^k+5^k$ is a square of an integer number

Answer 1

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.

Answer 2

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?

Answer 3

Even values of $k \rightarrow \equiv 2 \bmod 8$. Odd values of $k \rightarrow \equiv 2 \bmod 3$.

Answer 4

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).

Answer 5

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.

Answer 6

Hint : You can consider using congruences

Modulo 3 and modulo 4

Or

Modulo 8