I have tested for all primes less than $40,000$ and it seems that, if $p$ is a prime then $p^2+26$ is not. I would like to see a proof or a counterexample.
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1https://math.stackexchange.com/questions/150952/provingallprimesare1or1modulo6 – 1110101001 Mar 04 '17 at 21:32
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If $x$ is not divisible by $3$, then $x^2 \equiv 1\mod3$. So for the primes we have
$$p^2+26\equiv 0\mod3$$
Edit.
As user236182 noted, we should separately treat the case of $p = 3$, because in this case my reasons are not valid. For this prime we have $p^2+26 = 35$, which is not prime.
Andrei Kulunchakov
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For primes $p > 3$, 3 divides the quantity $p^2  1$ (proof provided here), so for some integer $n$, $$3n = p^2  1$$ and $$p^2 + 26 = (p^2  1) + 27 = 3n + 27 = 3(n+9)$$ Since 3 does not divide any primes greater than 3, and by observation $p^2 + 26$ is not prime for $p \le 3$, we can conclude that $p^2 + 26$ is composite.
infinitylord
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