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Subtle Summed Squares (Posted on 2007-05-18) Difficulty: 3 of 5
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.

See The Solution Submitted by Old Original Oskar!    
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Solution Solution | Comment 2 of 6 |

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
  Posted by K Sengupta on 2007-05-18 14:55:29

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