Suppose a number N can be written as P times Q times R times..., where all of P, Q, R... can each be written as the sum of two perfect nonzero squares.
Show that in this case N itself can also be written as the sum of two perfect squares.
We denote N = A_1*A_2*.... for simplicity rather than
N = P*Q*R.......
Let us denote, N_t = Prod(j=1 to t) A_j, where A_j= p_j^2 + q_j^2
for all j = 1,2,...,t
Now, N_1*N_2
= (p_1^2+q_1^2)((p_2^2+q_2^2)
= A^2 + B^2, where:
(A, B) = [(p_1*p_2 +q_1*q_2), |p_1*q_2 - p_2*q_1|]
Or, [|p_1*p_2 -q_1*q_2|, (p_1*q_2 + p_2*q_1)]
Thus, the result is true for t=2
Let the result be true for t=s
We know that A_(s+1) = p_(s+1)^2 + q_(s+1)^2
Letting N_s = C^2 + D^2, it can be easily proved as before that:
N_(s+1) = E^2 + F^2, where
(E, F) = (C*p_(s+1)+ D*q_(s+1)), |C*q_(s+1) - D*p_(s+1)|]
Or, [|C*p_(s+1)- D*q_(s+1)|, (C*q_(s+1) + D*p_(s+1)]
Thus, the result is also true for t=s+1
Consequently, the product N will always be expressible as the sum of two squares.
Edited on May 18, 2007, 2:59 pm
Solution
