11
$\begingroup$

The question is asking us to prove that $x^2+y^2+z^2=999$ has no integer solutions.

Attempt at a solution:

So I've noticed that since 999 is odd, either one of the variables or all three of the variables must be odd.

If I assume that only one variable is odd, I can label the variables like this: $$x=2k_1+1$$ $$y=2k_2$$ $$z=2k_3$$

By substituting, and doing some algebra, I can conclude that $k_1^2+k_2^2+k_3^2+k_1=249.5$, which is not possible since all $k_i\in\Bbb Z$.

If all three are odd, I can rename the variables like this: $$x=2k_1+1$$ $$y=2k_2+1$$ $$z=2k_3+1$$ Eventually I conclude that $k_1^2+k_2^2+k_3^2+k_1+k_2+k_3 = 249$, but I don't know where to go from there.

An alternative I've considered is brute-forcing it, but I'd rather avoid that if I can. Any assistance here would be greatly appreciated.

  • 4
    Welcome to the site and nice first question! Well-tagged, well-formatted, includes what you've tried. I wish we could frame it! This isn't really my area, but I suspect an approach like the first will work for the second case, but it's just a hunch. I think the numbers must have other properties as well that give your contradiction. – 2017-01-26
  • 4
    Are you done with $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{1}+k_{2}+k_{3}=249$, rewriting this as $k_{1}(k_{1}+1)+k_{2}(k_{2}+1)+k_{3}(k_{3}+1)=249$, where each term is even? – 2017-01-26
  • 0
    https://en.m.wikipedia.org/wiki/Legendre's_three-square_theorem – 2017-01-26
  • 1
    @Jan you should make that comment an answer. – 2017-01-26
  • 0
    @Jan: Wow! It's so simple; I can't believe I missed that! Thank you! – 2017-01-26

3 Answers 3

4

This is impossible because the number is congruent to $-1\bmod 8$.

Notice that squares are only $1,4$ and $0\bmod 8$.

In fact there is a theorem by Legendre that say that a number is not the sum of three squares if and only if it is of the form $4^a(8b-1)$. (the other direction is the tough one).

  • 1
    there is an accessible proof in Dickson, Modern Elementary Theory of Numbers (1939). Just matrices, inequalities, what have you. – 2017-01-26
  • 0
    Is it possible to do with minkowski's theorem? that is the proof I learned of the sum of two squares theorem. – 2017-01-26
  • 1
    Probably. There is a different approach in Serre's little book, that an integer is the sum of three integer squares if and only if it is the sum of the squares of three rational numbers. Fairly strong property. – 2017-01-26
  • 1
    http://alpha.math.uga.edu/~pete/Clark_Jagy_11_13_2013.pdf – 2017-01-26
  • 0
    Nice, I didn't know that you and Peter were co-authors. – 2017-01-26
10

Using congruences . . .

Odd squares are always $1 \pmod 8$, hence also $1 \pmod 4$.

Even squares are always $0 \pmod 4$, hence either $0 \text{ or } 4 \pmod 8$.

Since $x^2 + y^2 + z^2$ is odd, either $x,y,z$ are all odd, or exactly one of $x,y,z$ is odd.

If $x,y,z$ are all odd, then $x^2 + y^2 + z^2 \equiv 3 \pmod 8$, contradiction, since $999 \equiv 7 \pmod 8$.

If exactly one of $x,y,z$ is odd, then $x^2 + y^2 + z^2 \equiv 1 \pmod 4$, contradiction, since $999 \equiv 3 \pmod 4$.

5

My immediate solution was the same as Jorge Fernández Hidalgo, using $\bmod 8$ limits, but carrying on from your sticking point (and trusting your work to that point):

$$k_1^2+k_2^2+k_3^2+k_1+k_2+k_3 = 249 \\ (k_1^2+k_1) + (k_2^2+k_2)+(k_3^2+k_3) = 249 \\ k_1(k_1+1) + k_2(k_2+1)+k_3(k_3+1) = 249 \\ $$ and we have three even terms summing to an odd number, which cannot therefore exist.