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I am stumped on this problem. The furthest I have got is to assume that there exists a largest number $N$ that cannot be written as the sum of $6$ squares. Ideally, I was trying to get a contradiction by constructing a larger number, dependant on $N$, that cannot be written as the sum of $6$ squares, but I have gotten nowhere with that.

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even if you don't allow 0 it's not true. Legendre's theorem says you can write a number as the sum of three squares iff it isn't in the form $4^n(8k+7)$.for n write it as the sum of $m+t$ such that m is sum of three nonzero square and t is of the form $8k+3$ which must be sum of three nonzero square. the only problem occurs when $n=8k+10=12+(8(k-1)+6)$ and again each of $12$ and $8k+6$ are sum of three nonzero square.

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