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USAJMO 2023 (Problem 1)

Find all triples of positive integers

(x,y,z)

that satisfy the equation

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

.

Core

Toolkit 26 — Expansion (or factorization) of three binomials

Major

Toolkit 24 — Square of a sum (n terms)

By Toolkit 24 — Square of a sum (n terms),

2(x+y+z+2xyz)^2

=2\left(x^2+y^2+z^2+4x^2y^2z^2+2xy+2xz+4x^2yz+2yz+4xy^2z+4xyz^2\right)

=2x^2+2y^2+2z^2+8x^2y^2z^2+4xy+4xz+8x^2yz+4yz+8xy^2z+8xyz^2

By Toolkit 24 — Square of a sum (n terms),

(2xy+2yz+2zx+1)^2

=4x^2y^2+4y^2z^2+4z^2x^2+1+8xy^2z+8xyz^2+8x^2yz

\qquad+4xy+4yz+4zx

By Hints 1 and 2,

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

implies

8x^2y^2z^2-4x^2y^2-4x^2z^2-4y^2z^2+2x^2+2y^2+2z^2-1=2023

By Toolkit 26 — Expansion (or factorization) of three binomials and Hint 3,

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

Prime factorize

2023

:

2023=7\cdot17^2

Since

x,y,z

are positive integers, each of

2x^2-1,\qquad2y^2-1,\qquad2z^2-1

is a positive divisor of

7\cdot17^2

.

Therefore,

2x^2-1\in\{1,7,17,119,289,2023\}

2x^2\in\{2,8,18,120,290,2024\}

x^2\in\{1,4,9,60,145,1012\}

Since

x

is a positive integer,

x\in\{1,2,3\}

By Hint 8,

2x^2-1\in\{1,7,17\}.

Similarly,

2x^2-1,\ 2y^2-1,\ 2z^2-1\in\{1,7,17\}

Since

2023=7\cdot17\cdot17,

by Hint 9,

2x^2-1,\ 2y^2-1,\ 2z^2-1

must be a permutation of

\{7,17,17\}.

Therefore,

x,y,z

must be a permutation of

\{2,3,3\}.

(x,y,z)=(2,3,3),(3,2,3),(3,3,2)

AMC 10B Fall 2021 (Problem 22)
USAMO 2018 (Problem 1)
AMC 12A 2024 (Problem 15)

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