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Except for $\ p=2\ $ primes split into two disjoint classes, $\ p\equiv1\mod4\ $ and $\ p\equiv3\mod4.\ $ Squares respect this partition, odd prime $\ p=m^2+n^2\ \Leftrightarrow\ p\equiv1\mod4.\ $ On the other hand, triangles $\ \binom k2\ $ are oblivious to the $\mod4\ $ classification, as well as to the other classification $\mod6\ $ (each prime different from $2$ and $3$ is congruent to $1$ or $-1$).

Question   What is a simple characterization of primes of the form $\ p=\binom m2+\binom n2\ ?\ $ or, in a sense, there are no simple characterizations (?).

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    $\begingroup$ These primes are tabulated at oeis.org/A117048 $\endgroup$
    – Gerry Myerson
    Commented Jul 10, 2021 at 4:45
  • $\begingroup$ @GerryMyerson, thank you (again!). ***** I don't see any connection between $\ 6\cdot k\pm1\ $ and sums of two triangular numbers -- I feel disappointed; but is there any non-trivial connection anyway? $\endgroup$
    – Wlod AA
    Commented Jul 10, 2021 at 4:53

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$p={m\choose 2}+{n\choose 2}$ is equivalent to $8p+2=(2m-1)^2+(2n-1)^2$. On the other hand, if $8p+2$ is a sum of two squares, these squares must be both odd (this is seen modulo 4). So, applying the criterion for representability as a sums of two squares, we get that $p$ is a sum of two triangular numbers if and only if the prime factorization of $4p+1$ does not contain prime divisors $\equiv 3\pmod 4$ in odd power. I do not think that there is a more explicit characterization.

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  • $\begingroup$ Fedor, great again! **** I thought about adjusting one of the (many) proofs about $\ p=x^2+y^2\ $ to the given topic. $\endgroup$
    – Wlod AA
    Commented Jul 10, 2021 at 5:47

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