Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1976 USAMO Problems/Problem 3"
 
Line 16: Line 16:
 
{{alternate solutions}}
 
{{alternate solutions}}
  
−
Rewrite the equation into <math>a^2+b^2-a^2b^2=-c^2</math> or <math>a^2+b^2-2a^2b^2+a^2b^2 = -c^2</math>. The LHS of this equation can be written as the sum of two squares, i.e., <math>(a-b)^2+(ab)^2 = -c^2</math>. It is easy to see that LHS <math>\leq 0</math> and RHS <math>\geq 0</math>. Thus, both sides must be equal to zero: <math>(a-b)^2+(ab)^2 = 0</math> and <math>c^2 = 0</math>. From there we conclude that <math>a=b=c=0</math> is the only solution.
+
 
 
== See Also ==
 
== See Also ==
 
{{USAMO box|year=1976|num-b=2|num-a=4}}
 
{{USAMO box|year=1976|num-b=2|num-a=4}}

Latest revision as of 23:00, 9 October 2020

Problem

Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$.

Solution

Either $a^2=0$ or $a^2>0$. If $a^2=0$, then $b^2=c^2=0$. Symmetry applies for $b$ as well. If $a^2,b^2\neq 0$, then $c^2\neq 0$. Now we look at $a^2\bmod{4}$:

$a^2\equiv 0\bmod{4}$: Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let $a=2a_1$, $b=2b_1$, and $c=2c_1$. Thus $a_1^2+b_1^2+c_1^2=4a_1^2b_1^2$. Since the LHS is divisible by four, all the variables are divisible by 4, and we must do this over and over again, and from infinite descent, there are no non-zero solutions when $a^2\equiv 0\bmod{4}$.

$a^2\equiv 1\bmod{4}$: Since $b^2\neq 0\bmod{4}$, $b^2\equiv 1\bmod{4}$, and $2+c^2\equiv 1\bmod{4}$. But for this to be true, $c^2\equiv 3\bmod{4}$, which is an impossibility. Thus there are no non-zero solutions when $a^2\equiv 1\bmod{4}$.

Thus the only solution is the solution above: $(a,b,c)=0$.



Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


See Also

1976 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1976_USAMO_Problems/Problem_3&oldid=134862"