Can you use a formula for writing products of two sums of squares as a sum of squares to create Pythagorean triplets from pairs of primitive Pythagorean triplets?
First, the formula for writing products of two sums of squares as a sum of squares is:
$(a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2$
Attempt:
A Pythagorean triplet is: $a^2+b^2=c^2$.
Two examples of a Pythagorean triplet is $(3,4,5)$ and $(5,12,13)$.
Where $3^2+4^2=5^2$ and $5^2+12^2=13^2$.
Now, using the previously mentioned formula:
$(3^2+4^2)(5^2+12^2)=(3 \cdot 12 - 4 \cdot 5)^2+(3 \cdot 5+4 \cdot 12)^2=5^2\cdot 13^2=4225$
So, $16^2+63^2=65^2$. Therefore the answer is yes. Is this sufficient? I can't help but feel I have overlooked something. The question thankfully doesn't say "prove", but maybe a "proof" is what I am looking for. Can anyone help me with this?