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Suppose $x$, $y$, and $z$ are integers that satisfy the system equations :

$x^2y$+ $y^2z$ + $z^2x$ $= 2186$

$y^2x$+ $z^2y$ + $x^2z$ $= 2188$

What is $x^2+y^2+z^2$ ?

Gerry Myerson
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mortimer-shaun
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1 Answers1

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Subtracting the first equation from the second, we have: $$ y^2(x - z) + z^2 (y-x) + x^2(z-y) = 2$$ $$ \implies \qquad (x - z)(y-x) (y - z) = 2 \tag {3}$$

Now considering the possible integer factors of 2, could you solve for possible $(x, y, z)$?

ADDITIONAL HINTS

a) The difference between these integers can be only among $\pm1, \pm2$, and should satisfy (3). So assume $x=a$ and $x <y<z$, can you write $y$ and $z$ in terms of $a$? What if the ordering is different - are there other solutions?

b) To find $a$, substitute solutions obtained from above into your first equation.

Macavity
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  • i've tried but i didn't find that it's equal to down vote favorite

    Suppose x, y, and z are integers that satisfy the system equations :

    $x2y+ y2z + z2x =2186$

    $y2x+ z2y + x2z =2188$

     – mortimer-shaun Apr 16 '13 at 03:34
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    Not sure what you are trying to say! Follow the path suggested, you will get a form for $(x, y, z)$ with only one parameter. Substitute into any equation, you will get the value for the parameter and hence the answer needed. Do try yourself. – Macavity Apr 16 '13 at 03:41
  • @RossMillikan Thanks! – Macavity Apr 16 '13 at 03:43
  • I don't see how the first equation in the (3) implies the second. – Gerry Myerson Apr 16 '13 at 05:41
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    @GerryMyerson You could expand and factorise the LHS, or expand both LHS and note they are the same, or simply note that $x=y$ makes the first LHS zero, hence $(x-y)$ is a factor. Similarly for the other factors, and the constant can be determined by checking the coefficient of any power, say of $x^2y$. – Macavity Apr 16 '13 at 05:46
  • @Mcavity how did you figure out how to factor like that when you solved this problem? Did you see that x=y is a solution? – Ovi Apr 22 '13 at 21:37
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    @Ovi A curious thing about the problem was that the cyclic sums were only 2 apart. So I hoped for a factorisation of 2, and it turned out to be simple. Not many generic strategies for Diophantine problems, unfortunately. – Macavity Apr 23 '13 at 02:14