USAJMO 2023 (Problem 1)Find all triples of positive integers (x,y,z)(x, y, z)(x,y,z) that satisfy the equation 2(x+y+z+2xyz)2=(2xy+2yz+2zx+1)2+20232(x + y + z + 2xyz)^2 = (2xy + 2yz + 2zx + 1)^2 + 20232(x+y+z+2xyz)2=(2xy+2yz+2zx+1)2+2023.Related TopicsToolkit 24 — Square of a sum (n terms)Toolkit 26 — Expansion (or factorization) of three binomialsHints (12)Hint 1Use 24. Squares of a sum.Hint 2Expand and simplify it.Hint 38x2y2z2−4x2y2−4y2z2−4z2x2+2x2+2y2+2z2−1=20238x^2y^2z^2 - 4x^2y^2 - 4y^2z^2 - 4z^2x^2 + 2x^2 + 2y^2 + 2z^2 - 1 = 20238x2y2z2−4x2y2−4y2z2−4z2x2+2x2+2y2+2z2−1=2023Hint 4Use 26. Factorization of three binomials.Hint 5(2x2−1)(2y2−1)(2z2−1)=2023(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023(2x2−1)(2y2−1)(2z2−1)=2023Hint 6Prime factorize 2023.Hint 7(2x2−1)(2y2−1)(2z2−1)=7×172(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 7 \times 17^2(2x2−1)(2y2−1)(2z2−1)=7×172Hint 8When a∈Z+a \in \mathbb{Z}^+a∈Z+, then 2a2−1∈{1,7,17,119,289,2023}2a^2 - 1 \in \{1, 7, 17, 119, 289, 2023\}2a2−1∈{1,7,17,119,289,2023}Hint 9a2=1,4,9,60,145,1012⇒a=1,2,3,×,×,×a^2 = 1, 4, 9, 60, 145, 1012 \Rightarrow a = 1, 2, 3, \times, \times, \timesa2=1,4,9,60,145,1012⇒a=1,2,3,×,×,×Hint 102x2−1,2y2−1,2z2−1∈{1,7,17}2x^2 - 1, 2y^2 - 1, 2z^2 - 1 \in \{1, 7, 17\}2x2−1,2y2−1,2z2−1∈{1,7,17}Hint 112023=7×17×172023 = 7 \times 17 \times 172023=7×17×17Hint 122x2−1,2y2−1,2z2−12x^2 - 1, 2y^2 - 1, 2z^2 - 12x2−1,2y2−1,2z2−1 is a permutation of {7,17,17}\{7, 17, 17\}{7,17,17}Final Answer(x,y,z)=(2,3,3),(3,2,3),(3,3,2)(x, y, z) = (2, 3, 3), (3, 2, 3), (3, 3, 2)(x,y,z)=(2,3,3),(3,2,3),(3,3,2)