1. ## Sum of squares

Prove that there are infinitely many positive integers $\displaystyle n$ such that $\displaystyle n(n + 1)$ can be expressed as a sum of two positive squares in at least two different ways. (Here $\displaystyle a^2 + b^2$ and $\displaystyle b^2 + a^2$ are considered as the same representation.)

2. Originally Posted by alexmahone
Prove that there are infinitely many positive integers $\displaystyle n$ such that $\displaystyle n(n + 1)$ can be expressed as a sum of two positive squares in at least two different ways. (Here $\displaystyle a^2 + b^2$ and $\displaystyle b^2 + a^2$ are considered as the same representation.)
Hint: If $\displaystyle A = a^2+b^2$ and $\displaystyle B=c^2+d^2$ then $\displaystyle AB = (ac-bd)^2 + (ad+bc)^2$.

3. Hello, Alex!

Prove that there are infinitely many positive integers $\displaystyle n$ such that $\displaystyle n(n + 1)$
can be expressed as a sum of two positive squares in at least two different ways.
(Here $\displaystyle a^2 + b^2$ and $\displaystyle b^2 + a^2$ are considered as the same representation.)

ThePerfectHacker hinted at an interesting identity,
. . one I had run across many years ago.

$\displaystyle (a^2+b^2)(c^2+d^2) \;=\;\begin{Bmatrix}(ac-bd)^2 + (ad + bc)^2 \\ (ac+bd)^2 + (ad-bc)^2 \end{Bmatrix}$

Let $\displaystyle c^2 \:=\:a^2+b^2$
. . and there are an infinite number of Pythagorean triples.

And we have: .$\displaystyle c^2(c^2+d^2)$

Let $\displaystyle d = 1$ and we have: .$\displaystyle c^2(c^2+1)$ which has the form: $\displaystyle n(n+1)$

Example: .use $\displaystyle 3^2+4^2\:=\:5^2\quad \begin{bmatrix} a\:=\:3 \\ b\:=\:4 \\ c\:=\:5 \\ d\:=\:1\end{bmatrix}$

$\displaystyle (3^2+4^2)(5^2+1^2) \;=\;\begin{Bmatrix}(3\cdot5-4\cdot1) + (3\cdot1 + 4\cdot5)^2 \\(3\cdot5 + 4\cdot1) + (3\cdot1 - 4\cdot5)^2 \end{Bmatrix}$

. . . . . .$\displaystyle (25)(26) \;=\;\begin{Bmatrix}11^2 + 23^2 \\ 19^2 + 17^2\end{Bmatrix}$ . . . all equal to 650

Example: use $\displaystyle 5^2+12^2 \:=\:13^2\quad\begin{bmatrix}a \:=\: 5 \\ b \:=\: 12 \\ c \:= \:13 \\ d \:=\: 1\end{bmatrix}$

$\displaystyle (5^2+12^2)(13^2+1^2) \;=\;\begin{Bmatrix}(5\cdot13 - 12\cdot1)^2 + (5\cdot1 + 12\cdot13)^2 \\ (5\cdot13 + 12\cdot1)^2 + (5\cdot1 - 12\cdot13)^2 \end{Bmatrix}$

. . . . . . $\displaystyle (169)(170) \;=\;\begin{Bmatrix}53^2 + 161^2 \\ 77^2 + 151^2\end{Bmatrix}$ . . . all equal to 28,730