Difference between revisions of "2022 USAMO Problems/Problem 4"
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+ | ==Solution 3== | ||
+ | |||
+ | We claim that the only solution is <math>(p,q) = (3,2)</math>. First, this clearly works, since <math>3-2=1^2</math> and <math>3\cdot2-2=2^2</math>. | ||
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+ | Start as in Solution 2 - assign positive integers <math>m, n,</math> such that <math>p-q=m^2</math> and <math>pq-q=n^2.</math> Then, <math>pq-p=n^2-m^2,</math> or <math>p(q-1)=(n-m)(n+m).</math> | ||
+ | |||
+ | First, <math>p=q</math> is impossible, since <math>p^2-p = p(p-1)</math> is not a perfect square. Now notice that <math>m<p</math> and <math>n<\max(p,q-1).</math> But <math>p-q>0,</math> so <math>p\geq q-1.</math> Now, <math>m, n < p.</math> | ||
+ | |||
+ | Therefore, <math>n-m</math> can't ever be a factor of <math>p,</math> and <math>n+m<2p,</math> so <math>n+m=p.</math> Then, <math>n-m=q-1,</math> so <math>p-q=2m-1.</math> But <math>p-q=m^2,</math> so <math>m^2=2m-1,</math> or <math>m=1.</math> Therefore, one of <math>p</math> and <math>q</math> must be <math>2</math> (and <math>1</math> is not a prime). | ||
+ | |||
+ | Thus, we conclude that our claim is true. | ||
==See also== | ==See also== | ||
{{USAMO newbox|year=2022|num-b=3|num-a=5}} | {{USAMO newbox|year=2022|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:58, 19 July 2025
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
Therefore,
or
. For each case,
. But
, this gives a contradiction. No solution for
.
We conclude that the only solution is .
(Lokman GÖKÇE)
Solution 3
We claim that the only solution is . First, this clearly works, since
and
.
Start as in Solution 2 - assign positive integers such that
and
Then,
or
First, is impossible, since
is not a perfect square. Now notice that
and
But
so
Now,
Therefore, can't ever be a factor of
and
so
Then,
so
But
so
or
Therefore, one of
and
must be
(and
is not a prime).
Thus, we conclude that our claim is true.
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.