It is an easy question by words, but not easy by solution! ☺

For any positive integer prime number $p$, Show that, for any positive integer $n>1$, $$n^p+p:\text{is not a perfect square.}$$

Can anyone help me too find a solution for it?

Thanks in advance!

P.S. For the case, $p=2$, (i.e. $n^2+2$: is not a perfect prime.) it is a classical exercise in elementary number theory. And for the case $p=7$, is a nice problem, (i.e. $n^7+7$: is not a perfect prime.) someone said it was a proposed problem to IMO.


closed as off-topic by Crostul, Pragabhava, user91500, user133281, user223391 Oct 28 '16 at 16:51

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  • 9
    $\begingroup$ Ah, a teaser....yet it is a false claim, of course: $\;1^3+3=2^2\;$ . Do I get some prize? $\endgroup$ – DonAntonio Oct 27 '16 at 15:10
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    $\begingroup$ @DonAntonio is the question wrong then? $\endgroup$ – tatan Oct 27 '16 at 15:17
  • $\begingroup$ @tatan Something unclear with my (counter) example? First, that is not a question but a teaser, as the OP seems to "know the solution" but, for some reason, wants to prove us. Second, the claim in the non-question is wrong, as shown. $\endgroup$ – DonAntonio Oct 27 '16 at 15:19
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    $\begingroup$ What if $n>1$? In any case I don't think this is an easy problem. $\endgroup$ – Xam Oct 27 '16 at 15:21
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    $\begingroup$ Mmm the case $p=2$ is easy because we have $n^2+2=k^2$, and the case $p=7$ was proposed in USA TST for IMO 2008. $\endgroup$ – Xam Oct 27 '16 at 15:23