Difference between revisions of "2023 USAJMO Problems/Problem 1"

Revision as of 17:58, 6 October 2023

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.

-FFAmplify

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.

@@ Line 39: / Line 39: @@
 Proceed as above.
+==See Also==
+{{USAJMO newbox|year=2023|before=First Question|num-a=2}}
+{{MAA Notice}}

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Difference between revisions of "2023 USAJMO Problems/Problem 1"

Revision as of 17:58, 6 October 2023

Problem

Solution 1

See Also