sum of squares of two rational numbers

[logistic_guy]

You have to count all divisors, not just the primes. But the formula says that it only applies if all [imath] s_j [/imath] are even which is not the case here. We have an odd [imath] s_1=1 [/imath] so [imath] r_2(120)=0. [/imath] My argument was wrong, I made a mistake.

 

[fresh_42]

\(\displaystyle 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\)

so

\(\displaystyle 1,5\) is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)
\(\displaystyle 3,15\) is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\)
so
\(\displaystyle d_1(n) - d_3(n) = 2 - 2 = 0\)

is it correct?

They say that the formula requires all [imath] s_j [/imath] to be even, so we cannot use it here.

Maybe it is correct, or it is just a matter of chance. This would require a look at the proof to decide whether the formula still holds in that case.