Difference between revisions of "1976 USAMO Problems/Problem 3"

Revision as of 14:05, 17 September 2012

Problem

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

Solution

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

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: $(a,b,c)=0$ .

@@ Line 13: / Line 13: @@
 Thus the only solution is the solution above: <math>(a,b,c)=0</math>.
-== See also ==
+== See Also ==
 {{USAMO box|year=1976|num-b=2|num-a=4}}
 [[Category:Olympiad Number Theory Problems]]

1976 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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Difference between revisions of "1976 USAMO Problems/Problem 3"

Revision as of 14:05, 17 September 2012

Problem

Solution

See Also