2023 USAJMO Problems/Problem 1: Difference between revisions
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==Solution 1== | |||
We claim that the only solutions are <math>(2,3,3)</math> and its permutations. | |||
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> | |||
Jumping on the coefficients in front of the <math>x^2</math>, <math>y^2</math>, <math>z^2</math> terms, we factor into: | |||
<math>(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023</math> | |||
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 | |||
Alternatively, a more obvious factorization is: | |||
<math>2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023</math> | |||
<math>(\sqrt{2}x+\sqrt{2}y+\sqrt{2}z+2\sqrt{2}xyz)^2-(2xy+2yz+2zx+1)^2=2023</math> | |||
<math>(2\sqrt{2}xyz+2xy+2yz+2zx+\sqrt{2}x+\sqrt{2}y+\sqrt{2}z+1)(2\sqrt{2}xyz-2xy-2yz-2zx+\sqrt{2}x+\sqrt{2}y+\sqrt{2}z-1)=2023</math> | |||
<math>(\sqrt{2}x+1)(\sqrt{2}y+1)(\sqrt{2}z+1)(\sqrt{2}x-1)(\sqrt{2}y-1)(\sqrt{2}z-1)=2023</math> | |||
<math>(2x^2-1)(2y^2-1)(2z^2-1)=2023</math> | |||
Proceed as above. ~eevee9406 | |||
note: idk gang i think the first factorization is more readily apparent but the second factorization is very orz indeed | |||
~Aarush12 | |||
==See Also== | |||
{{USAJMO newbox|year=2023|before=First Question|num-a=2}} | |||
{{MAA Notice}} | |||
Latest revision as of 23:18, 17 March 2025
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
note: idk gang i think the first factorization is more readily apparent but the second factorization is very orz indeed ~Aarush12
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
