Big sum of squares

Number Theory Level 5

4 475 137 = 201 6 2 + 64 1 2 4 627 633 = 197 7 2 + 84 8 2 \large 4\:475\:137 = 2016^2+641^2 \\ \large 4\:627\:633 = 1977^2+848^2 The two large numbers above are prime numbers . Consider their product, N = 4 475 137 × 4 627 633 = 20 709 291 660 721 N = 4\:475\:137 \times 4\:627\:633 = 20\:709\:291\:660\:721

Consider all possible ways in which N N can be written as the sum of two squares: N = a 2 + b 2 , a ≥ b ≥ 1. N = a^2 + b^2,\ \ \ a \geq b \geq 1.

What is the average value of the possible values of a a ? If you think N N cannot be written as the sum of two squares, type 999.


The answer is 3985632.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

In general, ( u 2 + v 2 ) ( x 2 + y 2 ) = ( u x ± v y ) 2 + ( u y ∓ v x ) 2 . (u^2 + v^2)(x^2 + y^2) = (ux \pm vy)^2 + (uy \mp vx)^2. If the factors are prime, there is no other way except these two to write the product as a sum of squares. Thus we find a 1 = 2016 ⋅ 1977 + 641 ⋅ 848 = 4 529 200 a 2 = 2016 ⋅ 1977 − 641 ⋅ 848 = 3 442 064 b 1 = 2016 ⋅ 848 − 641 ⋅ 1977 = 442 311 b 2 = 2016 ⋅ 848 + 641 ⋅ 1977 = 2 976 825 \begin{array}{ll} a_1 = 2016\cdot 1977 + 641\cdot 848 = 4\:529\:200 & a_2 = 2016\cdot 1977 - 641\cdot 848 = 3\:442\:064 \\ b_1 = 2016\cdot 848 - 641 \cdot 1977 = 442\:311 & b_2 = 2016\cdot 848 + 641 \cdot 1977 = 2\:976\:825 \end{array}

The average of the two a a values is 2016 ⋅ 1977 = 3 985 632 . 2016 \cdot 1977 = \boxed{3\:985\:632}.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice problem and clear solution! In the first formula there is a small issue with the letters.

Otto Bretscher - 5 years, 2 months ago

Same solution.Such a long problem with such an easy solution.BRILLIANT!!!!

rajdeep brahma - 4 years, 2 months ago

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