Prove there is infinite number of 3 consecutive numbers which are sum of 2 squares.

Question

Prove there is infinite number of 3 consecutive numbers which are sum of 2 squares of ($0\notin $) natural numbers. Example: $$72 = 6^2+6^2$$ $$73= 8^2+3^2$$ $$74 =7^2+5^2$$

I could only find 232,233 and 234 is next such triple. I was also thinking about to implement $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc) ^2$$ but don't know how.

Answer 1

If we have a solution where $ n-1 = a^2 + b^2, n = c^2 + d^2, n+1 = e^2 + f^2 $, with all variables non-zero, then

• $n$ has to be odd, because $\{n-1, n, n+1\} \equiv \{0, 1, 2\} \pmod{4}$.
• $n^2 -1 = (ae+bf)^2 + (af - be)^2$ -> If either of these are 0, then $n^2 = 1 + m^2$ which easily leads to a contradiction.
• $ n^2 = (c^2 - d^2)^2 + (2cd)^2$. -> Since $n$ is odd, so $ c \neq d$ and thus $c^2 - d^2 \neq 0$.
• $n^2 + 1 = n^2 + 1$.

Thus, we have our 3 consecutive integers

So, starting with your seed of $ n = 73$, an infinite family exists.

Answer 2

A more efficient but more messy construction. Let $n = 50a^2 + 14a$. Then

$$(3n)^2 + (4n)^2, \\ (5n)^2 + 1^2, \\ (3n + 8a + 1)^2 + (4n - 6a - 1)^2\\ $$

are consecutive sums of two nonzero squares.