2023 USAJMO Problems/Problem 1: Difference between revisions
Factorization 2 |
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-FFAmplify | -FFAmplify | ||
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. | |||
Revision as of 17:52, 6 October 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.
-FFAmplify
Alternatively, a more obvious factorization is:
Proceed as above.
