Jean-Paul's Email  Message from July 3, 2013

Dear Doron

I have just seen your nice paper "Generalizing and Implementing
Michael Hirschhorn?s AMAZING Algorithm for Proving Ramanujan-Type
Congruences" on ArXiv.

About your elementary lemma on page 7: I guess that a thousand people
already wrote you to provide a proof :-))
If not (or not yet :-), here is my 1c contribution (where epsilon more
is proved).

With my very best wishes
jean-paul

>>>>>
Let \ell be a prime congruent to 3 modulo 4. Let r \equiv (\ell - 6)/24.
Then if n_1 and n_2 in [0, \ell) satisfy
n_1(n_1+1)/2 + n_2(n_2+1)/2  \equiv r modulo \ell
then BOTH n_1 and n_2 are equal to (\ell - 1)/2.

Proof. Write n_1(n_1+1) = (n_1 + 1/2)^2 - 1/4, and analogously for n_2.
        We thus have
        [(n_1 + 1/2)^2]/2 - 1/8 + [(n_2 + 1/2)^2]/2 - 1/8 \equiv r modulo \ell
        i.e.,
        [(n_1 + 1/2)^2]/2 + [(n_2 + 1/2)^2]/2 \equiv \ell/24 \equiv 0  
modulo \ell
        hence
        (n_1 + 1/2)^2 + (n_2 + 1/2)^2 \equiv 0 \modulo \ell
        Letting x_i := (n_i + 1/2), we have x_1^2 + x_2^2 \equiv 0 modulo \ell.
        If x_1 is not zero modulo \ell, then x_2/x_1 is a square root of -1
        modulo \ell which is impossible (\ell \equiv 3 mod 4), thus  
x_1 \equiv 0.
        Hence also x_2 \equiv 0 modulo \ell.
        In other words n_1 and n_2 are equivalent to -1/2 modulo \ell, i.e.,
        to (\ell - 1)/2 modulo \ell.
>>>>>