# Math contest integer triplet problem

Can any one help me with this?

Determine all integer triples (x,y,z) such that 1 ≤ x ≤ y ≤ z and x + y + z + xy + yz + xz = xyz − 1.

I thought of Vietta's formula but don't let me lead you into a deadend.

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Wonderful ideas and thanks for the help. I encourage others to post other approaches as well. But could you guys try to elaborate each step as clearly as possible? I'm nowhere near the proficiency you all are. Much thanks – Yadnarav3 Aug 31 '13 at 0:31
Thanks to everyone for their help – Yadnarav3 Aug 31 '13 at 0:58

Presumably there is a clever way, since this is a contest problem. The following approach is unpleasantly unclever, and the details are somewhat tedious.

Note first that $x$ cannot be $1$, for if it is the left-hand side is bigger than the right-hand side.

Note also that $x\lt 4$. Suppose to the contrary that all our numbers are $\ge 4$. Divide both sides by $xyz$. Then the terms on the left have sum $\le \frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{15}{16}$, while the right is $\ge 1-\frac{1}{64}$.

So $x=2$ or $x=3$. We end up with two accessible problems in $2$ variables.

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The equation can be rewritten as

$$(1+x)(1+y)(1+z)=2xyz$$

and this rewrites as

$$\left(1+{1\over x}\right)\left(1+{1\over y}\right)\left(1+{1\over z}\right)=2$$

This already shows that all three numbers are greater than $1$, since otherwise the product on the left would be greater than $2$. Since we're assuming $x$ is the smallest of the three positive integers, we must have

$$1+{1\over x} \ge 2^{1/3}\approx 1.2599$$

hence $x=2$ or $3$. If we let $x=2$, the same idea shows that

$${4\over3}\gt 1+{1\over y} \ge \sqrt{4\over3}\approx1.1547$$

which implies $4\le y\le6$, and if we let $x=3$, it gives

$${3\over2}\gt1+{1\over y} \ge \sqrt{3\over2}\approx1.2247$$

which implies $3\le y \le 4$. Out of all this, we get the triples

$$(2,4,15), (2,5,9), (2,6,7), (3,3,8),\text{ and } (3,4,5)$$

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Thank you for the help, I got this one the most – Yadnarav3 Aug 31 '13 at 0:49
@user87611, good, I'm glad it helped. – Barry Cipra Aug 31 '13 at 0:57

You can rewrite the equation as $2(x+y+z)=(x-1)(y-1)(z-1)$. Since the right side is cubic and the right is quadratic, the variables must be small or the right will be too big. Let us set $u=x-1,v=y-1,w=z-1$ for convenience, making the equation $2(u+v+w)+6=uvw$ Starting from small $u$, if $u=1$ we get $2(v+w)+8=uv, (u-2)(v-2)=12$, leading to solutions $(2,4,15),(2,5,9),(2,6,7)$. If $u=2$ we get $2(v+w)=2vw, 24=(2v-2)(2w-2)$ leading to $(3,3,8),(3,4,5)$. As André Nicolas has shown, that is it.

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What led you to rewriting it that way? – Yadnarav3 Aug 30 '13 at 23:59
I looked for ways to factor the expression, or some large part of it. The symmetry of it all made me look at the $(x-1)(y-1)(z-1)$ term, which got the cubic and constant terms and see how it accounted for most of it. When it also did the quadratics it seemed likely I was onto something. – Ross Millikan Aug 31 '13 at 0:16

Your formula leads to $(x+1)(y+1)(z+1) = 2xyz$. Note none of $x,y,z$ equal $1$. Also if all $x,y,z$ are greater than or equal to $4$, then $(x+1)(y+1)(z+1) \leq (125/64)xyz < 2xyz$, a contradiction. Similarly assuming $x = 2$ or $x = 3$, then you cannot have $y$ and $z$ both greater than or equal to $7$, because otherwise $(x+1)(y+1)(z+1) \leq (3/2)x(y+1)(z+1) \leq (3/2)(64/49)xyz < 2xyz$. So in the worst case, you only need to check a small number of possible values of $x,y$ to see if there is integer solution for $z$. If you assume $x = 2,3$ and $2 \leq y \leq 6$, the remaining solutions follow by symmetry.

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I don get dem!!! – Yadnarav3 Aug 31 '13 at 0:27