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I am confused as how to solve an equation with three squared variables to get its integer solutions? As: $$x^2+y^2+z^2=200$$ Thanks!

Edward Jiang
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John
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    There are only finitely many cases to check. Trying them yourself you find that $\pm 6, \pm 8, \pm 10$ are solutions. – Hubble Dec 27 '14 at 05:44

3 Answers3

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All of $x^2$, $y^2$, and $z^2$ must be non-negative, so there are only finitely many cases to check.

Edward Jiang
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If you are doing it to find solution to your previous question about the area of that triangle then you should use heron's formula and simplify it in terms of $x, y, z$ like,

$\sqrt{s(s-a)(s-b)(s-c)}=\dfrac{1}{4}\sqrt{(x+y+z)(x+y-z)(x-y+z)(-x+y+z)}$, which will simplify into,

$\dfrac{1}{4}\sqrt{-(x^4+y^4+z^4)+2(x^2y^2+y^2z^2+z^2x^2)}$

Now you can substitute the values to get the required area.

Dheeraj Kumar
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  • Thanks for your reply, my issue was mainly with how to get the values of x, y and z. I knew this could be solved by limiting the value of any variable to a proper interval and some trial-and-error but thought there might be some way to get the solution directly. – John Dec 27 '14 at 05:51
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I would do it this way: (it takes just few minutes)

Let's assume $x<=y<=z$, without loss of generality.

$z$ must be larger than $8$, since $3*8^2 = 192 < 200$.

$z$ must be smaller than $15$, since $15^2 = 225 > 200$.

Now, there are six possibilities:

$$x^2+y^2=200-9^2=119$$ $$x^2+y^2=200-10^2=100$$ $$x^2+y^2=200-11^2=79$$ $$x^2+y^2=200-12^2=56$$ $$x^2+y^2=200-13^2=31$$ $$x^2+y^2=200-14^2=4$$

Then, according to the list of Pythagorean triples, only this solution can exist:

$$x=6, y=8, z=10$$

The list above assumes x<>0, and for x=0, we easily get two more solutions:

$$x=0, y=2, z=14$$

and

$$x=0, y=10, z=10$$

Of course, if variables can be negative, there is also a number of solutions directly derived from last three ones.

VividD
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