USA TST 2013 #2

by Wolstenholme, Oct 17, 2014, 2:05 AM

Find all triples $(x,y,z)$ of positive integers such that $x \leq y \leq z$ and
$x^3(y^3+z^3)=2012(xyz+2).$

Solution:

Lemma 1: $503 \nmid x(y^2 - yz + z^2)$

Proof: Note that $503 \vert x \Longrightarrow 503^3 \vert 2012(xyz + 2) \Longrightarrow 503 \vert (4xyz + 8)$ but clearly $4xyz + 8 \equiv 8 \pmod{503}$ , contradiction! Moreover, $503 \vert y^2 - yz + z^2 \Longrightarrow 503 \vert (2y - z)^2 + 3z^2 \Longrightarrow \left(\frac{-3}{503}\right) = 1$ . Since $503 \equiv -1 \pmod{4} \Longrightarrow \left(\frac{-1}{503}\right) = -1$ , this means that $\left(\frac{3}{503}\right) = -1$ . But by Quadratic Reciprocity we have $\left(\frac{3}{503}\right)\left(\frac{503}{3}\right) = -1$ and since $\left(\frac{503}{3}\right) = \left(\frac{2}{3}\right) = -1$ , we know that $\left(\frac{3}{503}\right) = 1$ , contradiction!

Lemma 2: $x \vert 2$

Proof: Note that $x^3(y^3 + z^3) = 2012(xyz + 2) \Longrightarrow x \vert 4012$ but since $(503, x) = 1$ by Lemma 1, this actually implies that $x \vert 8$ . But if $4 \vert x$ , then $v_2(2012(xyz + 2)) = 3 < v_2(x^3(y^3 + z^3))$ , contradiction!

Note that Lemma 1 implies that $503 \vert (y + z)$ . Now, we proceed with casework.

Case 1: $x = 1$

Our equation becomes $(y + z)(y^2 - yz + z^2) = 2012(yz + 2)$ . Since the RHS is even, we must have $y \equiv z \pmod{2}$ . Moreover, since by AM-GM we have $(y + z)(y^2 - yz + z^2) \ge (y + z)yz$ it is clear that $y + z < 2515$ . This means that either $y + z = 1006$ or $y + z = 2012$ .

If $y + z = 2012$ then the equation can be written as $(y - z)^2 = 2$ which clearly has no integer solution. If $y + z = 1006$ then the equation becomes $y^2 - 3yz + z^2 = 4$ . But since in this case $y + z = 1006 \Longrightarrow y^2 + 2yz + z^2 = 1006^2$ this means that $5yz = 1006^2 - 4$ which also clearly has no integer solution.

Case 2: $x = 2$

Our equation becomes $(y + z)(y^2 - yz + z^2) = 503(yz + 1)$ . Since by AM-GM $(y + z)(y^2 - yz + z^2) \ge (y + z)yz$ it's clear that we must have $y + z = 503$ . Then the equation becomes $(y - z)^2 = 1$ and since we are given that $y \le z$ we obtain the unique solution $\boxed{(x, y, z) = (2, 251, 252)}$ .

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
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by fukano_2, Jun 14, 2020, 6:24 AM
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by AopsUser101, Jan 29, 2020, 8:27 PM
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by mathleticguyyy, Jun 5, 2019, 8:26 PM
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by User360702, Jan 10, 2019, 6:03 PM
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by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
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by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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USA TST 2013 #2

by Wolstenholme, Oct 17, 2014, 2:05 AM

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