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Difference between revisions of "2022 USAMO Problems/Problem 4"

(Created page with "==Solution== Since <math>q(p-1)</math> is a perfect square and <math>q</math> is prime, we should have <math>p - 1 = qb^2</math> for some positive integer <math>b</math>. Let...")
 
(→‎Solution 2)
 
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==Solution==
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==Problem==
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Find all pairs of primes <math>(p, q)</math> for which <math>p-q</math> and <math>pq-q</math> are both perfect squares.
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==Solution 1==
  
 
Since <math>q(p-1)</math> is a perfect square and <math>q</math> is prime, we should have <math>p - 1 = qb^2</math> for some positive integer <math>b</math>. Let <math>a^2 = p - q</math>. Therefore, <math>q = p - a^2</math>, and substituting that into the <math>p - 1 = qb^2</math> and solving for <math>p</math> gives
 
Since <math>q(p-1)</math> is a perfect square and <math>q</math> is prime, we should have <math>p - 1 = qb^2</math> for some positive integer <math>b</math>. Let <math>a^2 = p - q</math>. Therefore, <math>q = p - a^2</math>, and substituting that into the <math>p - 1 = qb^2</math> and solving for <math>p</math> gives
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Thus, the only solution is <math>(p, q) = (3, 2)</math>.
 
Thus, the only solution is <math>(p, q) = (3, 2)</math>.
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==Solution 2==
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Let <math>p-q = a^2</math>, <math>pq - q = b^2</math>, where <math>a, b</math> are positive integers. <math>b^2 - a^2 = pq - q - (p-q) = pq -p</math>. So,
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<cmath> b^2 - a^2 = p(q-1) \tag{1}</cmath>
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<math>\bullet</math> For <math>q=2</math>, <math>p = b^2 - a^2 = (b-a)(b+a)</math>. Then <math>b-a=1</math> and <math>b+a=p</math>. <math>a=\dfrac{p-1}{2}</math> and <math>p-q = a^2</math>. Thus, <math>p - 2 = \left( \dfrac{p-1}{2} \right)^2 \implies p^2 - 6p + 9 = 0</math> and we find <math>p=3</math>. Hence <math>(p,q) = (3,2)</math>.
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<math>\bullet</math> For <math>q=4k+3</math>, (<math>k\geq 0</math> integer), by <math>(1)</math>, <math>p(4k+2) = b^2 - a^2</math>. Let's examine in <math>\mod 4</math>, <math>b^2 - a^2 \equiv 2 \pmod{4}</math>. But we know that <math>b^2 - a^2 \equiv 0, 1 \text{ or } 3 \pmod{4}</math>. This is a contradiction and no solution for <math>q = 4k + 3</math>.
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<math>\bullet</math> For <math>q=4k+1</math>, (<math>k > 0</math> integer), by <math>(1)</math>, <math>p(4k) = b^2 - a^2</math>. Let <math>k=m\cdot n</math>, where <math>m\geq n \geq 1</math> and <math>m, n</math> are integers. Since <math>p>q</math>, we see <math>p>4k</math>. Thus, by <math>(1)</math>, <math> (b-a)(b+a) = 4p\cdot m \cdot n</math>. <math>b-a</math> and <math>b+a</math> are same parity and  <math>4p\cdot m \cdot n</math> is even integer. So, <math>b-a</math> and <math>b+a</math> are both even integers. Therefore,
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<math>
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\left\{ \begin{array}{rcr}
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b+a =  & 2pn \
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b-a =  & 2m
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\end{array} \right.
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</math>
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or
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<math>
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\left\{ \begin{array}{rcr}
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b+a =  & 2pm \
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b-a =  & 2n
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\end{array} \right.
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</math>
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Therefore, <math>a=pn - m</math> or <math>a = pm - n</math>. For each case, <math>p-q = p - 4mn - 1 < a</math>. But <math>p-q = a^2</math>, this gives a contradiction. No solution for <math>q = 4k + 1</math>.
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We conclude that the only solution is <math>(p,q) = (3,2)</math>.
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(Lokman GÖKÇE)
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==See also==
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{{USAMO newbox|year=2022|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 18:43, 10 April 2024

Problem

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

Solution 1

Since $q(p-1)$ is a perfect square and $q$ is prime, we should have $p - 1 = qb^2$ for some positive integer $b$. Let $a^2 = p - q$. Therefore, $q = p - a^2$, and substituting that into the $p - 1 = qb^2$ and solving for $p$ gives \[p = \frac{a^2b^2 - 1}{b^2 - 1} = \frac{(ab - 1)(ab + 1)}{b^2 - 1}.\] Notice that we also have \[p = \frac{a^2b^2 - 1}{b^2 - 1} = a^2 + \frac{a^2 - 1}{b^2 - 1}\] and so $b^2 - 1 | a^2 - 1$. We run through the cases

  • $a = 1$: Then $p - q = 1$ so $(p, q) = (3, 2)$, which works.
  • $a = b$: This means $p = a^2 + 1$, so $q = 1$, a contradiction.
  • $a > b$: This means that $b^2 - 1 < ab - 1$. Since $b^2 - 1$ can be split up into two factors $F_1, F_2$ such that $F_1 | ab - 1$ and $F_2 | ab + 1$, we get

\[p = \frac{ab - 1}{F_1} \cdot \frac{ab + 1}{F_2}\] and each factor is greater than $1$, contradicting the primality of $p$.

Thus, the only solution is $(p, q) = (3, 2)$.

Solution 2

Let $p-q = a^2$, $pq - q = b^2$, where $a, b$ are positive integers. $b^2 - a^2 = pq - q - (p-q) = pq -p$. So, \[b^2 - a^2 = p(q-1) \tag{1}\]

$\bullet$ For $q=2$, $p = b^2 - a^2 = (b-a)(b+a)$. Then $b-a=1$ and $b+a=p$. $a=\dfrac{p-1}{2}$ and $p-q = a^2$. Thus, $p - 2 = \left( \dfrac{p-1}{2} \right)^2 \implies p^2 - 6p + 9 = 0$ and we find $p=3$. Hence $(p,q) = (3,2)$.


$\bullet$ For $q=4k+3$, ($k\geq 0$ integer), by $(1)$, $p(4k+2) = b^2 - a^2$. Let's examine in $\mod 4$, $b^2 - a^2 \equiv 2 \pmod{4}$. But we know that $b^2 - a^2 \equiv 0, 1 \text{ or } 3 \pmod{4}$. This is a contradiction and no solution for $q = 4k + 3$.


$\bullet$ For $q=4k+1$, ($k > 0$ integer), by $(1)$, $p(4k) = b^2 - a^2$. Let $k=m\cdot n$, where $m\geq n \geq 1$ and $m, n$ are integers. Since $p>q$, we see $p>4k$. Thus, by $(1)$, $(b-a)(b+a) = 4p\cdot m \cdot n$. $b-a$ and $b+a$ are same parity and $4p\cdot m \cdot n$ is even integer. So, $b-a$ and $b+a$ are both even integers. Therefore,

$\left\{ \begin{array}{rcr} b+a = & 2pn \\ b-a = & 2m \end{array} \right.$ or $\left\{ \begin{array}{rcr} b+a = & 2pm \\ b-a = & 2n \end{array} \right.$ Therefore, $a=pn - m$ or $a = pm - n$. For each case, $p-q = p - 4mn - 1 < a$. But $p-q = a^2$, this gives a contradiction. No solution for $q = 4k + 1$.


We conclude that the only solution is $(p,q) = (3,2)$.

(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

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