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

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==Solution 1==
 
==Solution 1==
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We first consider the case where one of <math>p,q</math> is even. If <math>p</math> equals 2, <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 \rightarrow (p,q)=(3,2)</math>.
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We first consider the case where one of <math>p,q</math> is even. If <math>p</math> equals 2, <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)=(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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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</math> <math>\rightarrow p(q-1)=(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.
  
  

Revision as of 13:21, 3 August 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 $p$ equals 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 $(pq-q)-(p-q)=y^2-x^2 \rightarrow$ $\rightarrow p(q-1)=(y+x)(y-x)$. 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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