From square to square

Number Theory Level 4

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

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

The answer is 319.

2 solutions

Arjen Vreugdenhil
May 2, 2016

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

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

Is this representation unique? Why?

Calvin Lin Staff - 5 years, 1 month ago

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

I checked.

$84\:421$ is a prime of the form $p = 4n + 1$ . 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,

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

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

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

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

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

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

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

very nice question..

Ramiel To-ong - 4 years ago

From square to square

The answer is 319.

2 solutions

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