Reach for the Summit - M-S1-A1

Algebra Level 2

Let set A = { x ∣ x = a 2 + b 2 , a , b ∈ Z } A=\{x|x=a^2+b^2, a,b \in \mathbb Z\} .

If x 1 , x 2 ∈ A x_1,x_2 \in A , is it always true that x 1 ⋅ x 2 ∈ A x_1 \cdot x_2 \in A ?

Reach for the Summit problem set - Mathematics

Yes \textup{Yes} No \textup{No}

Alice Smith by Alice Smith , 20, China

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2 solutions

David Vreken
Jul 14, 2020

If x 1 = a 1 2 + b 1 2 x_1 = a_1^2 + b_1^2 and x 2 = a 2 2 + b 2 2 x_2 = a_2^2 + b_2^2 ,

then x 1 ⋅ x 2 x_1 \cdot x_2

= ( a 1 2 + b 1 2 ) ( a 2 2 + b 2 2 ) = (a_1^2 + b_1^2)(a_2^2 + b_2^2)

= a 1 2 a 2 2 + b 1 2 b 2 2 + a 1 2 b 2 2 + a 2 2 b 1 2 = a_1^2a_2^2 + b_1^2b_2^2 + a_1^2b_2^2 + a_2^2b_1^2

= a 1 2 a 2 2 + 2 a 1 a 2 b 1 b 2 + b 1 2 b 2 2 + a 1 2 b 2 2 − 2 a 1 a 2 b 1 b 2 + a 2 2 b 1 2 = a_1^2a_2^2 + 2a_1a_2b_1b_2 + b_1^2b_2^2 + a_1^2b_2^2 - 2a_1a_2b_1b_2 + a_2^2b_1^2

= ( a 1 a 2 + b 1 b 2 ) 2 + ( a 1 b 2 − a 2 b 1 ) 2 ∈ A = (a_1a_2 + b_1b_2)^2 + (a_1b_2 - a_2b_1)^2 \in A

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Applying Fermat's theorem on the sum of two squares we get the result.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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