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Difference between revisions of "2023 USAJMO Problems/Problem 1"
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==Problem==
 
Find all triples of positive integers <math>(x,y,z)</math> that satisfy the equation
 
Find all triples of positive integers <math>(x,y,z)</math> that satisfy the equation
  
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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==Solution 1==
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We claim that the only solutions are <math>(2,3,3)</math> and its permutations.
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Factoring the above squares and canceling the terms gives you:
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<math>8(xyz)^2 + 2(x^2 +y^2 + z^2) = 4((xy)^2 + (yz)^2 + (zx)^2) + 2024</math>
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Jumping on the coefficients in front of the <math>x^2</math>, <math>y^2</math>, <math>z^2</math> terms, we factor into:
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<math>(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023</math>
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Realizing that the only factors of 2023 that could be expressed as <math>(2x^2 - 1)</math> are <math>1</math>, <math>7</math>, and <math>17</math>, we simply find that the only solutions are <math>(2,3,3)</math> by inspection.
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-Max
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Alternatively, a more obvious factorization is:
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<math>2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023</math>
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<math>(\sqrt{2}x+\sqrt{2}y+\sqrt{2}z+2\sqrt{2}xyz)^2-(2xy+2yz+2zx+1)^2=2023</math>
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<math>(2\sqrt{2}xyz+2xy+2yz+2zx+\sqrt{2}x+\sqrt{2}y+\sqrt{2}z+1)(2\sqrt{2}xyz-2xy-2yz-2zx+\sqrt{2}x+\sqrt{2}y+\sqrt{2}z-1)=2023</math>
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<math>(\sqrt{2}x+1)(\sqrt{2}y+1)(\sqrt{2}z+1)(\sqrt{2}x-1)(\sqrt{2}y-1)(\sqrt{2}z-1)=2023</math>
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<math>(2x^2-1)(2y^2-1)(2z^2-1)=2023</math>
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Proceed as above. ~eevee9406
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==See Also==
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{{USAJMO newbox|year=2023|before=First Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 21:01, 28 February 2024

Problem

Find all triples of positive integers $(x,y,z)$ that satisfy the equation

\begin{align*} 2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023 \end{align*}


Solution 1

We claim that the only solutions are $(2,3,3)$ and its permutations.

Factoring the above squares and canceling the terms gives you:

$8(xyz)^2 + 2(x^2 +y^2 + z^2) = 4((xy)^2 + (yz)^2 + (zx)^2) + 2024$

Jumping on the coefficients in front of the $x^2$, $y^2$, $z^2$ terms, we factor into:

$(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023$

Realizing that the only factors of 2023 that could be expressed as $(2x^2 - 1)$ are $1$, $7$, and $17$, we simply find that the only solutions are $(2,3,3)$ by inspection.

-Max


Alternatively, a more obvious factorization is:

$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023$

$(\sqrt{2}x+\sqrt{2}y+\sqrt{2}z+2\sqrt{2}xyz)^2-(2xy+2yz+2zx+1)^2=2023$

$(2\sqrt{2}xyz+2xy+2yz+2zx+\sqrt{2}x+\sqrt{2}y+\sqrt{2}z+1)(2\sqrt{2}xyz-2xy-2yz-2zx+\sqrt{2}x+\sqrt{2}y+\sqrt{2}z-1)=2023$

$(\sqrt{2}x+1)(\sqrt{2}y+1)(\sqrt{2}z+1)(\sqrt{2}x-1)(\sqrt{2}y-1)(\sqrt{2}z-1)=2023$

$(2x^2-1)(2y^2-1)(2z^2-1)=2023$

Proceed as above. ~eevee9406

See Also

2023 USAJMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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