<s>[INST] Using formula of sum of two squares ... create Pythagorean triplets from pairs of primitive Pythagorean triplets? 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?
 [/INST] Let the following be your Pythagorean triplets.
$$a_1^2 +b_1^2 = c_1^2$$
$$a_2^2 +b_2^2 = c_2^2$$
We have (from the formula):
$$(a_1^2 +b_1^2)\cdot(a_2^2 +b_2^2)=(a_1\cdot b_2-a_2\cdot b_1)^2+(a_1\cdot a_2-b_1\cdot b_2)^2$$
Obviously the RHS is the sum of two squares.  To prove this is a Pythagorean triplet, we need to show the LHS is a perfect square.  We know that the sum $a_1^2 +b_1^2 = c_1^2$, and similar for the other factor.  Thus, we have:
$$(c_1^2)\cdot(c_2^2)=(a_1\cdot b_2-a_2\cdot b_1)^2+(a_1\cdot a_2-b_1\cdot b_2)^2$$
$$(c_1\cdot c_2)^2=(a_1\cdot b_2-a_2\cdot b_1)^2+(a_1\cdot a_2-b_1\cdot b_2)^2$$
Which is a Pythagorean triple.
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