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$0, 1, 2$ is an example of three consecutive non-negative integers $n, n+1, n+2$ which are each the sum of two integer squares. Using modular arithmetic you can prove that in all of these triplets $n \equiv 0 \pmod 8$ .

Now I'm wondering, is there a way to find all such triplets? Below are two ways I've found which generate infinite triplets, but not all:

Because $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$, you can take any triplet (except $0, 1, 2$ which will give itself) and generate a new triplet. If $n, n+1, n+2$ is the chosen triplet then the following is also a triplet:

$$n(n+2) = n^2 + 2n$$

$$(n+1)^2 + 0^2 = n^2 + 2n + 1$$

$$(n+1)^2 + 1^2 = n^2 + 2n + 2$$

Assume that there exists a triplet in the following form (letting $b^2 = 2a^2 + 1$):

$$a^2 + a^2 = 2a^2$$

$$2a^2 + 1 = b^2 + 0^2$$

$$2a^2 + 2 = b^2 + 1$$

Solving $b^2 = 2a^2 + 1$ using the convergents for $\sqrt 2$ you'll find that every second convergent ($3/2$, $17/12$, $99/70$, etc.) will be in the form $b/a$.

For example:

$$70^2 + 70^2 = 9800$$

$$99^2 + 0^2 = 9801$$

$$99^2 + 1^2 = 9802$$

Sp3000
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    The sequence of numbers n such that n, n+1, n+2 are all sums of two squares is given at http://oeis.org/A082982 . (As you'd expect, they're all multiples of 8.) This sequence doesn't look to me like there's a nice formula. Of course this doesn't prove there isn't one. – Michael Lugo Jun 20 '11 at 15:07
  • I believe you can find a complete solution. Here's one possible attack:
    1. Let $n = n_1^2+n_2^2$ for integers $n_1,n_2 \ge 0$.
    2. Rearrange the double relation $$n_1^2+n_2^2+2=n_3^2+n_4^2+1=n_5^2+n_6^2$$ so that you can [twice] apply Cauchy’s classical solution to the linear equation $$rX+sY+tZ=0.$$
    3. “Simplify”.

    Obviously, Step 3 is the hardest. ;)

     – Kieren MacMillan Aug 04 '16 at 16:53

2 Answers2

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The numbers $4n^4+4n^2,4n^4+4n^2+1,4n^4+4n^2+2$ are all sums of two squares, so there's another infinite family. I don't think you're going to find a useful formula that will give all solutions, but I don't see how to prove this.

Gerry Myerson
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Well we know $a^2 + b^2 = n$ so $a^2 + b^2 + 1 = c^2 + d^2$

A parameterization for this can be found here: https://mathoverflow.net/questions/104621/seeking-an-integer-parameterization-for-a2b2-c2d21 (just scroll to the bottom)

and then you just do this again after the substitutions: $c^2 + d^2 + 1 = g^2 + f^2$

Just find a way to set the c and d from both equations equal to each other while remaining in the integer domain and you should have a solid parameterization.

ServingSpy
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