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Are there any references to study the integer solutions (existence and how many) of Diophantine equations like $a^2+b^2=c^2+d^2+2$, $a^2+b^2=c^2+d^2+3$, $a^2+b^2=c^2+d^2+5$...? Actually, I can prove that there are integer solutions (a,b,c,d) for any integers n. But I don't know how to count them. Thanks a lot.

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  • $\begingroup$ You are essentially looking at numbers in this sequence at distance of 2 from each other. $\endgroup$
    – Wojowu
    Commented May 24, 2016 at 21:15
  • $\begingroup$ For any $n\in\mathbb{Z}$ the 4-tuples $(n\pm1,n\mp1,n,n)$ are a solution, so there are obviously infinitely many for this one (not sure what you mean by "like"). $\endgroup$ Commented May 24, 2016 at 21:30
  • $\begingroup$ Thanks, very nice a special solution. I have edited my questions, should be $a^2+b^2=c^2+d^2+n$ for any integer n. $\endgroup$ Commented May 24, 2016 at 21:34

2 Answers 2

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This is a special case of the shifted convolution problem for modular forms. For example, see Chapter 12 of Iwaniec's book "Spectral methods for automorphic forms" (AMS Grad studies in Math 53). Theorem 12.5 there gives (for $n$ odd) $$ \sum_{m\le x} r(m) r(m+n) = 8 \Big(\sum_{d|n} d^{-1} \Big) x + O(x^{2/3} n^{1/3+\epsilon}), $$ where $r(m)$ denotes the number of ways of writing $m$ as a sum of two squares. You'll find more references in the book.

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Here's an elementary way to see that there are always plenty of solutions. Find $r$ and $s$ such that $r+s=n$ and neither $r$ nor $s$ is twice an odd number --- for large $n$, there will be many ways to do this. Then, there always exist $a,b,c,d$ such that $a^2-c^2=r$ and $b^2-d^2=s$, whence $a^2+b^2=c^2+d^2+n$.

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