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Revision as of 11:11, 11 March 2024
Contents
Problem
Find all pairs of primes for which
and
are both perfect squares.
Solution 1
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 .
Solution 2
Let ,
, where
are positive integers.
. So,
For
,
. Then
and
.
and
. Thus,
and we find
. Hence
.
For
, (
integer), by
,
. Let's examine in
,
. But we know that
. This is a contradiction and no solution for
.
For
, (
integer), by
,
. Let
, where
and
are integers. Since
, we see
. Thus, by
,
.
and
are same parity and
is even integer. So,
and
are both even integers. Therefore,
or gay sex
Therefore,
or
. For each case,
. But
, this gives a contradiction. No solution for
.
We conclude that the only solution is .
(Lokman GÖKÇE)
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.