Let's learn how to find the integer solutions of a three variable equation.
Problem: Consider the following equation: \( (x-y)^2 + (y-z)^2 + (z - x)^2 = 2018 \). Find the integer solutions to this three variable equation.
Discussion: Set x - y = a, y - z = b. Then z - x = - (a+b). Clearly, we have, \( a^2 + b^2 + (-(a+b))^2 = 2018 \). Simplifying we have \( a^2 + b^2 + ab = 1009 \). Now, treating this as a quadratic in a, we have:
$$ a^2 + ba + b^2 - 1009 = 0 $$
Hence \( a = \frac{-b \pm \sqrt{b^2 - 4(b^2 - 1009)}}{2} = \frac{-b \pm \sqrt{4 \times 1009 - 3b^2}}{2} \)
Since a is an integer, we must have \( 4 \times 1009 - 3b^2 \) (the discriminant) to be a positive perfect square integer. This severely limits the number of possibilities for b. For example, we need \( b^2 \le \frac{4 \times 1009}{3} \) or \( b \le 36 \). So one may 'check' for these 36 values.
Only b = 35 works. Then \( a = \frac{-35 \pm \sqrt{4 \times 1009 - 3\times 35^2}}{2} \). But this makes \( a \) negative.
Reducing the number of cases:
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But the Question doesn't says it should be a positive integer??
So 35 can be the ans!
Yes. Once we restrict ourselves to positive solutions. It won't be difficult to track down the negative ones.