# 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?

• What you did works in general, same argument. Suppose that $a^2+b^2=s^2$, $c^2+d^2=t^2$. Then the identity you quoted gets us $x$, $y$ such that $x^2+y^2=(st)^2$. We need to show this is a Pytagorean triple, so to make sure that the $ad-bc\ne 0$. By the way, minor typo in what you wrote, it is $65^2$ not $64^2$. Dec 10 '12 at 2:47

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.