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.
>>>>>