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# Integer solutions of a three variable equation

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:

• Suppose b is the smaller of a and b (WLOG) then $1009 = a^2 + ab + b^2 \ge b^2 + b*b + b^2 = 3b^2$ or $1009/3 \ge b^2$ or 19 > b.
• Also, b is 0, 1 or -1 mod 7 (if $4 \times 1009 - 3*b^2$ needs to be a perfect square). Hence we bring it down to 6 cases: b = 6,7,8,13,14,15

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