From square to square

25 9 2 + 31 9 2 = 168 842 a 2 + b 2 = 84 421 \begin{array} { c c c r } 259^2 & + & 319^2 & = &168 \: 842 \\ a^2 & + & b^2 & = &84 \: 421 \\ \end{array}

Assuming a a and b b are positive integers, find a + b a + b .


The answer is 319.

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2 solutions

Note that 84 421 = 1 2 168 842 ; a 2 + b 2 = 1 2 ( 25 9 2 + 31 9 2 ) ; a 2 + b 2 = 1 2 ( ( a b ) 2 + ( a + b ) 2 ) . 84\:421 = \tfrac12\cdot 168\:842; \\ a^2 + b^2 = \tfrac12\left(259^2 + 319^2\right); \\ a^2 + b^2 = \tfrac12\left((a-b)^2 + (a+b)^2\right).

Thus we set a b = 259 a - b = 259 , a + b = 319 a+b = \boxed{319} . It is easy to check that a = 289 a = 289 and b = 30 b = 30 are indeed integers.

Is this representation unique? Why?

Calvin Lin Staff - 5 years, 1 month ago

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A very good question... of course, given the fact that Brilliant only accepts one answer, it should be unique.

I checked.

84 421 84\:421 is a prime of the form p = 4 n + 1 p = 4n + 1 . That guarantees that there exists a unique representation as the sum of squares.

Arjen Vreugdenhil - 5 years, 1 month ago
Manuel Kahayon
May 10, 2016

By the Fibonacci-Brahmagupta Identity,

( a 2 + b 2 ) ( c 2 + d 2 ) = ( a c + b d ) 2 + ( b c a d ) 2 (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(bc-ad)^2

Plugging in c = d = 1 c=d=1 , we get

( a 2 + b 2 ) ( 1 + 1 ) = ( a + b ) 2 + ( b a ) 2 (a^2+b^2)(1+1)=(a+b)^2+(b-a)^2

Now, we let ( a + b ) 2 + ( b a ) 2 = 168842 (a+b)^2+(b-a)^2=168842 to get

2 ( a 2 + b 2 ) = ( a + b ) 2 + ( b a ) 2 = 168842 2(a^2+b^2)=(a+b)^2+(b-a)^2=168842 or a 2 + b 2 = 84421 a^2+b^2=84421

Now, from the format above, we see that a + b = 319 a+b = 319 , b a = 259 b-a = 259

So, a + b = 319 a+b = \boxed{319}

very nice question..

Ramiel To-ong - 4 years ago

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