I would like to ask for a prove about an observation I did regarding the sums of squares and prime numbers (in another question here), or a counterexample of it.
My capabilities to do this kind of demonstrations are very poor, so I do not know if in this case this is difficult to prove or not... I would also appreciate very much some hints about it:
Definitions:
$S1_n, S2_n\ /$
$S1_n = n^2 + (n-1)^2$
$S2_n = n^2 + (n+1)^2$
$n \ge 5, n \in \Bbb N\ $
Observation to prove:
$(S1_n,S2_n) \in \Bbb P \ $ iif $\ n=5k , k \in \Bbb N$
Explanation: for values of $n$ greater or equal to $5$, only when $n$ is a multiple of 5 there is a possibility that both $S1$ and $S2$ were prime numbers. If $n$ is not a multiple of $5$ never is possible that $S1$ and $S2$ were at the same time prime numbers. Not all multiples of 5 generate a couple of primes, but only them seem to be able to generate those prime couples when $n$ is greater or equal to $5$.
Just in case: I have tested this using Python and I did not find any counterexamples in the interval $[5,299680]$, but I could be wrong. And the only example of a $n$ not being multiple of $5$ is $n=2$, $S1_2=5,S2_2=13$.
If there are questions to clarify I will update the text. Thank you!
You are right. If $n$ is not a multiple of $5$, then $n$ is congruent to $1$, $2$, $3$, or $4$ modulo $5$. If $n\equiv 1\pmod{5}$, then $n^2+(n+1)^2$ is divisible by $5$, for $n^2+(n+1)^2\equiv 1^2+2^2\equiv 0\pmod{5}$. If $n\equiv 2\pmod{5}$, then $n^2+(n-1)^2$ is divisible by $5$.
Similar considerations hold for $n\equiv 3\pmod{5}$ and $n\equiv 4\pmod{5}$. When $n\gt 2$, the number divisible by $5$ is greater than $5$, so cannot be prime.