Talk:Odd Numbers Not Expressible as Sum of 4 Distinct Non-Zero Coprime Squares

I wrote a little program to check this, and indeed you can't do this to $57, 65, 71$.

In fact I just found out if you express an odd number $\equiv 2 \pmod 3$ as sum of $4$ squares two of the squares must be divisible by $3$.

This fact directly follows from $n^2 \equiv 1 \pmod 3$ if $n \not \equiv 0 \pmod 3$, forcing $2 = 1 + 1 + 0 + 0$.

Thus this list is infinite.

I think the theorems Halter-Koch proved only concerns distinct non-zero squares (not coprime), including those for $5$ and above.

But Sum of Distinct Squares quoted the result as such, I don't want to make assumptions as I don't understand German. RandomUndergrad (talk) 20:54, 3 May 2020 (EDT)


 * Many thanks for the analysis of this. I'll take this further and see if I can puzzle out what that paper actually says -- my German however is limited to what I've picked up casually at work, and is very limited. IN any case yet another page needs to be added to the errata list of the Wells work. --prime mover (talk) 03:27, 4 May 2020 (EDT)


 * Further thoughts on this: the original statement in Wells actually says:


 * The largest odd integer which cannot be expressed as the sum of $4$ distinct non-zero squares with greatest common divisor $1$ is $157$.


 * Nothing about the squares needing to be mutually coprime. That was my mistake. --prime mover (talk) 06:57, 4 May 2020 (EDT)


 * At least (im falle $n \not \equiv 0 \bmod 8$ auch teilerfremden) makes sense now. RandomUndergrad (talk) 09:13, 4 May 2020 (EDT)