2
$\begingroup$

Given natural numbers $x,y,z$ such that it satisfies $x^{2}y+y^{2}z+z^{2}x-23=xy^{2}+yz^{2}+zx^{2}-25=3xyz$. Find the possible maximum value of $x+y+z$.

For the above problem I've tried using the identity $x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-xz-yz)$ but ultimately lead to nowhere... (same thing with AM-GM, but that just might be cause I didn't dig deep enough)

Is there any other identity I could use?

$\endgroup$
3
  • 1
    $\begingroup$ If you are restricting to natural numbers, analytic methods might not help much. The optimal integer solution might not be near the optimal real solution. I'd try to solve the simultaneous system in integers...I only see one solution (up to permutation), but I didn't try that hard. $\endgroup$
    – lulu
    Feb 12 at 12:38
  • 4
    $\begingroup$ Hint: $(x^2y+y^2z+z^2x) - (xy^2+yz^2+zx^2) = -(x-y)(y-z)(z-x)$ $\endgroup$ Feb 12 at 12:48
  • $\begingroup$ Please edit your post to credit the original source where you encountered this. See math.stackexchange.com/help/referencing $\endgroup$
    – D.W.
    Feb 19 at 18:06

1 Answer 1

3
$\begingroup$

Subtract the first two equalities: $$x^2(y-z)+y^2(z-x)+z^2(x-y)=-2$$

Now we can see that the polynomial on the left hand side easily factors, it can be written as $$(x-y)(y-z)(z-x)=2$$ Since $x,y,z \in \mathbb N$ we know each of these terms in the product can only be $\pm 2$ and $\pm 1$. WLOG let that $z>y>x$, and now we are just left to do a little casework.

Case $1$: $y-x=2$, $z-x=1$, $z-y=1$ has no solution.

Case $2$: $y-x=1$, $z-x=2$, $z-y=1$ has solution $(x,x+1,x+2)$.

Case $3$: $y-x=1$, $z-x=1$, $z-y=2$ has no solution.

So now we only need to solve for $x$ such that $$x^2(x+1)+(x+1)^2(x+2)+(x+2)^2x -23 = 3x(x+1)(x+2)$$ which after expanding becomes $x=7$.

Therefore, $(x,y,z)=(7,8,9)$ up to permutation and hence $$\boxed{x+y+z =24}$$

$\endgroup$
2
  • $\begingroup$ Ah, I see now. Thanks! $\endgroup$
    – JAB
    Feb 12 at 13:08
  • $\begingroup$ @JAB you're welcome $\endgroup$
    – Sahaj
    Feb 12 at 13:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .