# Does bounded version of Lagrange's 4-square theorem follow from this result?

Let $$S_4(n)$$ be the number of ways of representing $$n$$ as a sum of 4 integer squares and suppose I can give you a proof of the result that $$\sum_{i=1}^nS_4(i) \sim \frac{\pi^2}{2}n^2$$.

Is this result enough to deduce that beyond some $$N$$, every integer can be represented as the sum of 4 squares (given that we are "weakly" suggesting that $$\sum_{i=1}^nS_4(i) >> n$$)?

• It's not true that $S_4(n)\sim\frac{\pi^2}{2}n^2$. What is true is that $\sum_{k=1}^nS_4(k)\sim\frac{\pi^2}{2}n^2$. – Lord Shark the Unknown Dec 15 '18 at 16:16
• Ah yes, I meant that! Let me just update this. – Isky Mathews Dec 15 '18 at 17:12
• @LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual... – Isky Mathews Dec 15 '18 at 17:13
• You need more than that asymptotic. – Lord Shark the Unknown Dec 15 '18 at 19:13