Difference between revisions of "2023 USAJMO Problems/Problem 1"
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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: | ||
| + | |||
| + | <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. | ||
Revision as of 21:23, 24 March 2023
Problem
Find all triples of positive integers 
that satisfy the equation
Solution 1
We claim that the only solutions are 
and its permutations.
Factoring the above squares and canceling the terms gives you:
Jumping on the coefficients in front of the 
, 
, 
terms, we factor into:
Realizing that the only factors of 2023 that could be expressed as 
are 
, 
, and 
, we simply find that the only solutions are by inspection.
