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I'm learning a basic lesson of number theory and get stuck with this :

Find all positive integers $n$ and prime numbers $p$ such that $n^p+3^p$ is a perfect square.

I found that if $p \ge 3$, then $n$ is in the form $4k+1$ but I'm not sure if I am on the right way. Any help would be welcome:). Thank you.

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    $\begingroup$ Isn't "basic" and "Olympiad" a bit of a contradiction? :-) $\endgroup$
    – joriki
    Commented Oct 23, 2015 at 5:08
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    $\begingroup$ @joriki, no, it isn't. However, this one does not look basic to me. $\endgroup$
    – zhoraster
    Commented Oct 23, 2015 at 5:11
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    $\begingroup$ Hmm just deleted my answer after I realised all I had proved was that $n$ is of the form $4k+1$ and the square number must be even. This one is not basic. $\endgroup$
    – Ian Miller
    Commented Oct 23, 2015 at 9:51
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    $\begingroup$ A useful fact: if $p$ is prime and $gcd(a,b)=1$, then $gcd(\frac{a^p+b^p}{a+b},a+b)$ divides $p$. So we can consider the case where 3 divides n, and separately, $n+3=p^tA^2$ and $n^p+3^p=(n+3)p^tB^2$, where $t \in \{0,1\}$. $\endgroup$
    – Aravind
    Commented Oct 23, 2015 at 13:35
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    $\begingroup$ For $p=2$, the only solution is $3^2+4^2=5^2$. I think, for $p>3$, there are no solutions, but I do not have a proof yet. The only solution seems to be $n=4,p=2$ $\endgroup$
    – Peter
    Commented Oct 25, 2015 at 23:46

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