Difference between revisions of "2022 USAJMO Problems/Problem 5"

Latest revision as of 18:04, 6 October 2023

Problem

Find all pairs of primes $(p,q)$ for which $p-q$ and $pq-q$ are both perfect squares.

Solution 1

We first consider the case where one of $p,q$ is even. If $p=2$ , $p-q=0$ and $pq-q=2$ which doesn't satisfy the problem restraints. If $q=2$ , we can set $p-2=x^2$ and $2p-2=y^2$ giving us $p=y^2-x^2=(y+x)(y-x)$ . This forces $y-x=1$ so $p=2x+1\rightarrow 2x+1=x^2+2 \rightarrow x=1$ giving us the solution $(p,q)=(3,2)$ .

Now assume that $p,q$ are both odd primes. Set $p-q=x^2$ and $pq-q=y^2$ so $(pq-q)-(p-q)=y^2-x^2 \rightarrow p(q-1)$ $=(y+x)(y-x)$ . Since $y+x>y-x$ , $p | (x+y)$ . Note that $q-1$ is an even integer and since $y+x$ and $y-x$ have the same parity, they both must be even. Therefore, $x+y=pk$ for some positive even integer $k$ . On the other hand, $p>p-q=x^2 \rightarrow p>x$ and $p^2-p>pq-q=y^2 \rightarrow p>y$ . Therefore, $2p>x+y$ so $x+y=p$ , giving us a contradiction.

Therefore, the only solution to this problem is $(p,q)=(3,2)$ .

~BennettHuang

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 ~BennettHuang
+==See Also==
+{{USAJMO newbox|year=2022|num-b=4|num-a=6}}
+{{MAA Notice}}

2022 USAJMO (Problems • Resources)
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Difference between revisions of "2022 USAJMO Problems/Problem 5"

Latest revision as of 18:04, 6 October 2023

Problem

Solution 1

See Also