Difference between revisions of "2022 USAMO Problems/Problem 4"
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Revision as of 19:29, 31 March 2022
Problem
Solution
Since is a perfect square and
is prime, we should have
for some positive integer
. Let
. Therefore,
, and substituting that into the
and solving for
gives
Notice that we also have
and so
. We run through the cases
: Then
so
, which works.
: This means
, so
, a contradiction.
: This means that
. Since
can be split up into two factors
such that
and
, we get
and each factor is greater than
, contradicting the primality of
.
Thus, the only solution is .
See also
2022 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.