Bump! Bump!

Correct me if I'm wrong : Take an odd $k\geq 3$ $4 \mid 4n \mid n^k+(3n)^k$ also $4 \mid 2014^k$ and $3^k \equiv -1 \pmod{4}$ so the whole sum is $\equiv -1 \pmod{4}$ so it can't be a perfect square for every $k$ .

ComplexPhi wrote: Correct me if I'm wrong : Take an odd $k\geq 3$ $4 \mid 4n \mid n^k+(3n)^k$ also $4 \mid 2014^k$ and $3^k \equiv -1 \pmod{4}$ so the whole sum is $\equiv -1 \pmod{4}$ so it can't be a perfect square for every $k$ . You are right. Maybe I made a mistake in the translation... I'll check...

The correct statement is " $3^k+n^k+ (3n)^k+ 2014^k$ is a perfect square for some $k$", not for all $k...$ Thanks to dgrozev!

So could you restate the whole problem please... Thanks!

MathPanda1 wrote: So could you restate the whole problem please... Thanks! Here you are: Find the smallest positive integer $n,$ such that $3^k+n^k+ (3n)^k+ 2014^k$ is a perfect square for some natural number $k,$ but not a perfect cube, for all natural numbers $k.$

The answer is $n=2$. If $n=1$, then $3^k+n^k+ (3n)^k+ 2014^k=2 \times 3^k+2014^k+1 \equiv 0+1+1 \equiv 2 (mod 3)$ i.e. $2 \times 3^k+2014^k+1$ is never a perfect square i.e. $n=1$ does not work. If $n=2$, then $3^k+n^k+ (3n)^k+ 2014^k=2^k+3^k+6^k+2014^k$. $k=1$ implies $2^k+3^k+6^k+2014^k=2025=45^2$ i.e. $2^k+3^k+6^k+2014^k$ is a perfect square for some $k$. Now, take $mod 7$: We have that $2^k+3^k+6^k+2014^k$ is equivalent to $2,4,5,2,4,4,2,4,5,2,4,4,2,4,5,2,4,4,... (mod 7)$ However, a perfect cube is always equivalent to $-1,0$, or $1 (mod 7)$, so $2^k+3^k+6^k+2014^k$ is never a perfect cube i.e. $n=2$ works. Therefore, $n=2$ is the minimum.