sum of squares of two rational numbers

You have to count all divisors, not just the primes. But the formula says that it only applies if all sj s_j are even which is not the case here. We have an odd s1=1 s_1=1 so r2(120)=0. r_2(120)=0. My argument was wrong, I made a mistake.
😭
 
1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\displaystyle 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120

so

1,5\displaystyle 1,5 is congruent to 1\displaystyle 1 modulo 4\displaystyle 4
3,15\displaystyle 3,15 is congruent to 3\displaystyle 3 modulo 4\displaystyle 4
so
d1(n)−d3(n)=2−2=0\displaystyle d_1(n) - d_3(n) = 2 - 2 = 0

is it correct?😥
They say that the formula requires all sj s_j 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.
 
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