Difference between revisions of "2023 USAJMO Problems/Problem 1"
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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. | 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. | ||
| − | - | + | -Max |
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<math>(2x^2-1)(2y^2-1)(2z^2-1)=2023</math> | <math>(2x^2-1)(2y^2-1)(2z^2-1)=2023</math> | ||
| − | Proceed as above. | + | Proceed as above. ~eevee9406 |
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==See Also== | ==See Also== | ||
{{USAJMO newbox|year=2023|before=First Question|num-a=2}} | {{USAJMO newbox|year=2023|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:01, 28 February 2024
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.
-Max
Alternatively, a more obvious factorization is:
Proceed as above. ~eevee9406
See Also
| 2023 USAJMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
