1984 AHSME Problems/Problem 21
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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
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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