Difference between revisions of "2022 USAJMO Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
− | We first consider the case where one of <math>p,q</math> is even. If <math>p</math> | + | We first consider the case where one of <math>p,q</math> is even. If <math>p=2</math>, <math>p-q=0</math> and <math>pq-q=2</math> which doesn't satisfy the problem restraints. If <math>q=2</math>, we can set <math>p-2=x^2</math> and <math>2p-2=y^2</math> giving us <math>p=y^2-x^2=(y+x)(y-x)</math>. This forces <math>y-x=1</math> so <math>p=2x+1\rightarrow 2x+1=x^2+2 \rightarrow x=1</math> giving us the solution <math>(p,q)=(3,2)</math>. |
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+ | Now assume that <math>p,q</math> are both odd primes. Set <math>p-q=x^2</math> and <math>pq-q=y^2</math> so <math>(pq-q)-(p-q)=y^2-x^2 \rightarrow p(q-1)</math> <math>=(y+x)(y-x)</math>. Since <math>y+x>y-x</math>, <math>p | (x+y)</math>. Note that <math>q-1</math> is an even integer and since <math>y+x</math> and <math>y-x</math> have the same parity, they both must be even. Therefore, <math>x+y=pk</math> for some positive even integer <math>k</math>. On the other hand, <math>p>p-q=x^2 \rightarrow p>x</math> and <math>p^2-p>pq-q=y^2 \rightarrow p>y</math>. Therefore, <math>2p>x+y</math> so <math>x+y=p</math>, giving us a contradiction. | ||
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Therefore, the only solution to this problem is <math>(p,q)=(3,2)</math>. | Therefore, the only solution to this problem is <math>(p,q)=(3,2)</math>. | ||
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+ | ~BennettHuang | ||
+ | |||
+ | ==See Also== | ||
+ | {{USAJMO newbox|year=2022|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:04, 6 October 2023
Problem
Find all pairs of primes for which
and
are both perfect squares.
Solution 1
We first consider the case where one of is even. If
,
and
which doesn't satisfy the problem restraints. If
, we can set
and
giving us
. This forces
so
giving us the solution
.
Now assume that are both odd primes. Set
and
so
. Since
,
. Note that
is an even integer and since
and
have the same parity, they both must be even. Therefore,
for some positive even integer
. On the other hand,
and
. Therefore,
so
, giving us a contradiction.
Therefore, the only solution to this problem is .
~BennettHuang
See Also
2022 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.