Sum of squares

Number Theory Level 3

Let $a,b,c,d$ be non-negative integers such that $\left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { c }^{ 2 }+{ d }^{ 2 } \right) =2017$ .

What is the value of ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }$ ?

The answer is 2018.

2 solutions

Linkin Duck
Apr 3, 2017

Since 2017 is prime, $(1,2017)$ and $(2017,1)$ are the only integral solutions of the equation $x\times y=2017$ .

So, ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1+2017=\boxed { 2018 } .$

For the sake of completeness, we also have to show that such a solution exists, that is e.g.

$a^2 + b^2 = 1$

$c^2 + d^2 = 2017$

We can find, that such pairs of integers really exist: { (0, ±1) , (±9, ±44) } gives us the solutions (since we can interchange the pairs and the variables within a pair and choose the signs at 3 of the variables, therefore we have 64 solutions altogether (and 8 solutions if we are only considering (non-negative) whole numbers).

Zee Ell - 4 years, 2 months ago

It's a really nice completeness, I forgot to include that to my solution.

Linkin Duck - 4 years, 2 months ago

Edwin Gray
Apr 10, 2019

Since 2017 is a prime, we must have c^2 + d^2 = 1,; let c^2 = 1, and a^2 + b^2 = 2017, so a^2 + b^2 + c^2 + d^2 = 2018.

Sum of squares

The answer is 2018.

2 solutions

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