1984 AHSME Problems/Problem 21
Problem
The number of triples of positive integers which satisfy the simultaneous equations
is
Solution
We can factor the second equation to get , so we see that must be a factor of , and since this is prime, or . However, if , then , which is impossible for the field of positive integers. Therefore, for all possible solutions. Substituting this into the original equations gives
and
.
From the second equation, , and substituting this into the first equation yields , or . Factoring this gives , so or . Both of these yield integer solutions for , giving or , respectively. Therefore, the only solutions are and , yielding solutions, .
See Also
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Preceded by Problem 20 |
Followed by Problem 22 | |
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