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

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== Solution ==
 
== Solution ==
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We have the trivial solution, <math>(a,b,c)=(0,0,0)</math>. Now WLOG, let the variables be positive.
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Either <math>a^2=0</math> or <math>a^2>0</math>. If <math>a^2=0</math>, then <math>b^2=c^2=0</math>. Symmetry applies for <math>b</math> as well. If <math>a^2,b^2\neq 0</math>, then <math>c^2\neq 0</math>. Now we look at <math>a^2\bmod{4}</math>:
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* Case 1: <math>3|a</math>
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Thus the RHS is a multiple of 3, and <math>b</math> and <math>c</math> are also multiples of 3. Let <math>a=3a_1</math>, <math>b=3b_1</math>, and <math>c=3c_1</math>. Thus <math>a_1^2+b_1^2+c_1^2=9a_1^2b_1^2</math>. Thus the new variables are all multiples of 3, and we continue like this infinitely, and thus there are no solutions with <math>3|a</math>.
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<math>a^2\equiv 0\bmod{4}</math>: Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let <math>a=2a_1</math>, <math>b=2b_1</math>, and <math>c=2c_1</math>. Thus <math>a_1^2+b_1^2+c_1^2=4a_1^2b_1^2</math>. 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 <math>a^2\equiv 0\bmod{4}</math>.
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<math>a^2\equiv 1\bmod{4}</math>: Since <math>b^2\neq 0\bmod{4}</math>, <math>b^2\equiv 1\bmod{4}</math>, and <math>2+c^2\equiv 1\bmod{4}</math>. But for this to be true, <math>c^2\equiv 3\bmod{4}</math>, which is an impossibility. Thus there are no non-zero solutions when <math>a^2\equiv 1\bmod{4}</math>.
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Thus the only solution is the solution above: <math>(a,b,c)=0</math>.
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*Case 2: 3 is not a divisor of <math>a</math>.
 
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Thus <math>a^2\equiv b^2\equiv 1\bmod{3}</math>, but for <math>a^2+b^2+c^2</math> to be a quadratic residue, <math>c^2\equiv 1\bmod{3}</math>, and we have that a multiple of 3 equals something that isn't a multiple of 3, which is a contradiction.
 
  
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Thus after both cases, the only solution is the trivial solution stated above.
 
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== See also ==
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== See Also ==
 
{{USAMO box|year=1976|num-b=2|num-a=4}}
 
{{USAMO box|year=1976|num-b=2|num-a=4}}
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{{MAA Notice}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

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

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