Difference between revisions of "1991 AIME Problems/Problem 1"
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ab&=880 | ab&=880 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | After | + | After finding the [[prime factorization]] of <math>880=2^4\cdot5\cdot11</math>, it's easy to obtain the solutions <math>(a,b)=(16,55)</math>. Thus |
<cmath>x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}</cmath> | <cmath>x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}</cmath> | ||
Revision as of 20:55, 4 September 2019
Problem
Find 
if 
and 
are positive integers such that
Solution
Solution 1
Define 
and 
. Then 
and 
. Solving these two equations yields a quadratic: 
, which factors to 
. Either 
and 
or 
and 
. For the first case, it is easy to see that 
can be 
(or vice versa). In the second case, since all factors of 
must be 
, no two factors of can sum greater than 
, and so there are no integral solutions for . The solution is 
.
Solution 2
Since 
, this can be factored to 
. As 
and 
are integers, the possible sets for (ignoring cases where 
since it is symmetrical) are 
. The second equation factors to 
. The only set with a factor of 
is , and checking shows that it is our solution.
Solution 3
Let 
, 
then we get the equations
After finding the prime factorization of 
, it's easy to obtain the solutions 
. Thus
![\[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}\]](http://latex.artofproblemsolving.com/5/3/3/533aebef3f8520fd92f23fe7de569ffd07a2d777.png)
~ Nafer
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
