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Prove that if $ p $ is prime number $(p >3)$, then the number $p^2+2015$ is

multiple of $24 $?

Thank you for any help

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  • $\begingroup$ No problem, thanks for thanking me :-) $\endgroup$ – Gregory Grant May 30 '15 at 14:38
  • $\begingroup$ Do you have any specific $p$ in mind? $\endgroup$ – PyRulez May 30 '15 at 14:50
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    $\begingroup$ only what i cited : p greater than 3 $\endgroup$ – zeraoulia rafik May 30 '15 at 14:52
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All primes $p>3$ are of the form $6n\pm1$. But $$(6n\pm1)^2 +2015=12n(3n\pm1) +2016$$ is divisible by $24$ since one of $n$ and $3n\pm1$ is even, and $2016=84\cdot24$.

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Note that $-1 \equiv 2015 \pmod{24}$. So you can consider $p^2 -1$ instead of $p^2 + 2015$. For help with this see for example Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.

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Hint: you can show first that it is a multiple of 8, then that it is a multiple of 3.

Other method: you calculate the equation modulo 24 and for $p$ use all numbers with $\gcd(p,24)=1$ and $0<p<24$

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  • $\begingroup$ That doesn't really rise to the level of an answer, you should probably have posted this as a comment. $\endgroup$ – Gregory Grant May 30 '15 at 14:39
  • $\begingroup$ @GregoryGrant OP asks for help, and this gives some. $\endgroup$ – Mark Bennet May 30 '15 at 14:40
  • $\begingroup$ @GregoryGrant should i give a full proof instead? $\endgroup$ – supinf May 30 '15 at 14:41
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    $\begingroup$ @MarkBennet Sure, but help comes in two forms, comments and answers. This is a comment, as best as I can ascertain from the rules of the site $\endgroup$ – Gregory Grant May 30 '15 at 14:41
  • $\begingroup$ @supinf I don't think it has to be a full proof, there is not really a clear line between what constitutes an answer and what constitutes a comment. But I left my comment before you made your first edit when it was just the first sentence. That definitely was too minimal to be a answer. $\endgroup$ – Gregory Grant May 30 '15 at 14:43
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Hint: use the fact that $2016=84\times 24$ to say transform the question to whether $p^2+2016-1=2016+p^2-1$ is divisible by $24$.

Further hints follow, which suggest a slightly different route through which avoids dealing with cases.

Then any prime greater than $3$ is of the form $6n\pm1$ so that $p^2-1=36n^2\pm 12n=24n+12n(n\pm1)$

And we can then note:

Then $n(n\pm1)$ is the product of two successive integers, hence is even.

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