Difference between revisions of "2023 USAJMO Problems/Problem 1"
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| − | Find all triples of positive integers <math>(x,y,z)</math> that satisfy the equation\begin{align*} 2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023 \end{align*} | + | ==Problem== |
| + | Find all triples of positive integers <math>(x,y,z)</math> that satisfy the equation | ||
| + | |||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | 2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023 | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | |||
| + | |||
| + | ==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 | ||
| + | |||
| + | ==See Also== | ||
| + | {{USAJMO newbox|year=2023|before=First Question|num-a=2}} | ||
| + | {{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. 
