It's this type of question again...

The last digit of $k^2$ can be 0,1,4,5,6 or 9. The last digit of $5^n+8^n$ can be 1 or 9 when $n$ is even(otherwise the last digit is 3 or7). So let $n=2p$ . Then $(k+8^p)(k-8^p)=5^{2p}$ . Hence $2(8^p)$ must be a power of 5, which is impossible.

First of all, k is obviously odd. It is commonly known that the square of an odd number is ALWAYS 1 mod 8. So we get that k^2 ≡ 1 ≡ 5^n +8^n ≡ 5^n mod 8. This means that n is even. Using mod three we have k^2≡(-1)^n+(-1)^n mod 3. Since n is even (-1)^n=1. So k^2≡2 mod 3. Though it is also commonly known that no square is 2 mod 3 so the answer is 0.

Lemma: all odd squares, $k^2$ , are $\equiv 1\mod{8}$ Proof: Case 1: $k$ can be expressed in the form $4j + 1$ where $j$ is a non-negative integer. Squaring, we attain $(4j + 1)^2 = 16j^2 + 8j + 1 \equiv 1\mod{8}$ . Case 2: $k$ can be expressed in the form $4c + 3$ where $c$ is a non-negative integer. Squaring once again yields $(4c + 3)^2 = 16j^2 + 24j + 9 \equiv 1\mod{8}$ . With this lemma, we can now solve the problem. First note, that since $8^n$ is always even and $5^n$ is always odd, the LHS is odd so the RHS is odd. Now, since we have that the RHS $\equiv 1\mod{8}$ , we can take both sides $\mod{8}$ to get that $8^n + 5^n \equiv 5^n \equiv 1\mod{8}$ . Let's analyse the first few powers of $5\mod{8} : 5^0 \equiv 1\mod{8} ; 5^1 \equiv 5\mod{8} ; 5^2 \equiv 1\mod{8} ; 5^3 \equiv 1\mod{8}$ . Here we have a repetition, meaning if and only if $n$ is even, $5^n \equiv 1\mod{8}$ . Now, we can substitute $n = 2x$ where $x$ is a non-negative integer, and move the $8^{2x}$ term over to the RHS to yield $5^{2x} = k^2 - 8^{2x}$ . This is a difference of two squares, meaning we can factor the RHS to get $5^{2x} = (k - 8^x)(k + 8^x)$ . Since we have the obvious fact that powers of $5$ are only divisible by smaller or equal powers of $5 (5^0 = 1)$ , we can let $k + 8^x = 5^a$ and $k - 8^x = 5^b$ . Now we have a system of equations, and adding them together yields $2k = 5^a + 5^b ; k = \frac{5^a + 5^b}{2}$ . Substituting back into the equation, we have $\frac{5^a + 5^b}{2} + 8^x = 5^a$ . Now, note that $8^x = (2^x)(2^x)(2^x) = (2^x)^3 = 2^{3x}$ . Now, applying this fact, we get $\frac{5^a + 5^b}{2} + 2^{3x} = 5^a ; 5^a + 5^b + 2^{3x+1} = 2(5^a) ; 2^{3x + 1} = 5^a - 5^b$ . This is clearly a contradiction, as no power of $2$ is divisble by $5$ , and we are done.