<s>[INST] Use induction to prove a product of sums of squares is a sum of squares For any natural number $n\ge 1$, given pairs $(a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ of integer numbers, there exist integer number $c$ and $d$ such that
$$\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$$

My initial approach is
Base Case: $(a_1^2+b_1^2) = a_1^2+b_1^2$ which is true. (Although it is trivial)
Prove the statement is true when $n=2$: We have $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$$
(Thanks André Nicolas for pointing it out) 
So if $a,b,c,d$ are integers, $ac,bd,ad,bc$ are all integers and integers are closed under addition and subtraction. Hence $(ac-bd),(ad+bc)$ are integers. 
Inductive Hypothesis: $\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$ is true
Inductive Step: $$\prod_{i=1}^{n+1}(a_i^2+b_i^2) = \prod_{i=1}^{n}(a_i^2+b_i^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2)$$ 
Where $c$ and $d$ are integers. 
But when we apply $n=2$, we have  $(c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (e^2 + f^2)$ where $e$ and $f$ are integers.
Hence, by the principle of induction, the statement we needed to prove is true.
 [/INST] The following identity is useful:
$$(a^2+b^2)(x^2+y^2)=(ax\pm by)^2 + (ay \mp bx)^2.\qquad\qquad(\ast)$$
This identity is connected with the multiplication of complex numbers.  Note that 
$$(a+ib)(x+iy)=ax-by+i(ay+bx).$$ 
The identity $(\ast)$ can be thought of as asserting that the product of the norms of $a+ib$ and $x+iy$, namely $\sqrt{(x^2+y^2)(a^2+b^2)}$, is equal to  the norm of their product, namely $\sqrt{(ax-by)^2 +(ay+bx)^2}$.
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