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How many ordered triples (x,y,z) of integers are there such that ? Does the question have a geometric interpretation?

MIRB16  Mar 12, 2018
 #1
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\(\displaystyle x^{2}+y^{2}+z^{2}=49,\)

is the equation of a sphere radius 7, centered at the origin, so what you are looking for are points on its surface such that all three co-ordinates are integers.

I don't see any way through it other than by trying each integer value of one of the variables in turn.

Choosing z, it's largest value is 7, and that gets us 

\(\displaystyle x^{2}+y^{2}=0, \text{ from which we have } x=0\text{ and }y=0.\)

If z = 6,

\(\displaystyle x^{2}+y^{2}=13,\text{ leading to }x=\pm2, y=\pm3,\text{ or }x=\pm3,y=\pm2.\)

If z = 5,

\(\displaystyle x^{2}+y^{2}=24,\text{ and there are no integer solutions for that.}\)

So on through to z = -7.

Guest Mar 12, 2018

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