February 2024
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In the Math Corner of my February 2024 Diary I posted the following brainteaser from the 2023 USA Junior Mathematical Olympiad (USAJMO).
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
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• Solution
The solution involves some basic algebraic rock-breaking, then an insight. First let's break rocks.
The left-hand side expands to
2x² + 2y² +2z² + 8x²y²z² + 4xy + 4zx + 4yz +8x²yz + 8xy²z + 8xyz²
The right-hand side expands to
4x²y² + 4y²z² + 4z²x² + 1 + 8x²yz + 8xy²z + 8xyz² + 4xy + 4zx + 4yz
Cancelling and rearranging, that original equality reduces to
8x²y²z² − 4x²y² − 4y²z² − 4z²x² + 2x² + 2y² +2z² − 1 = 2023
That's enough rock-breaking. Here comes the insight.
If 2x² = 1 the left-hand side of that vanishes. Likewise if 2y² = 1 or 2z² = 1.
So now the original equality can be further reduced to
(2x² − 1)(2y² − 1)(2z² − 1) = 2023
Now we just need to know: Which three numbers, each of them one short of a square doubled, can be multiplied together to get 2023?
Since 2023 factorizes to 7 × 17² the following six numbers divide into it exactly: 1, 7, 17, 119, 289, 2023. Of those six, only 7 and 17 are one short of a square doubled: 7 is one short of 8, the double of 4, and 17 is one short of 18, the double of 9.
So 2x² − 1, 2y² − 1, and 2z² − 1 are 7, 17, and 17 in some order. Then x, y, and z must be 2, 3, and 3 in some order.
Answer: (2, 3, 3), (3, 2, 3), (3, 3, 2) are the only triples satisfying the original equality.