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

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Q. Find all pairs of primes

(p, q)

for which

p-q

and

pq-q

are both perfect squares.

Denote Perfect Squares: Let's denote the perfect squares as $x^2$ and $y^2$ , where $x$ and $y$ are integers. So we have: $\newline$ $p - q = x^2 \ldots(1)$ $\newline$ $pq - q = y^2 \ldots(2)$
Express $p$ in terms: From equation ( $1$ ), we can express $p$ in terms of $q$ and $x$ : $p = x^2 + q$ ...( $3$ )
Substitute $p$ into $q$ : Substitute $p$ from equation ( $3$ ) into equation ( $2$ ): $\newline$ $(x^2 + q)q - q = y^2$ $\newline$ $x^2q + q^2 - q = y^2$ $\newline$ $q(x^2 + q - 1) = y^2$ ...( $4$ )
Factor $y$ as $kq$ : Since $q$ is a prime number, and it divides the left side of the equation, it must also divide the right side. Therefore, $q$ must be a factor of $y^2$ . This implies that $y$ is a multiple of $q$ , so we can write $y$ as $y = kq$ , where $k$ is an integer.
Express $q$ in terms: Substitute $y = kq$ into equation ( $4$ ): $\newline$ $q(x^2 + q - 1) = (kq)^2$ $\newline$ $x^2 + q - 1 = k^2q$ $\newline$ $x^2 = k^2q - q + 1$ ...( $5$ )
New Perfect Square: Since $x^2$ is a perfect square and $k^2q - q + 1$ is an integer, $k^2q - q + 1$ must also be a perfect square. Let's denote this new perfect square as $z^2$ , so we have: $\newline$ $z^2 = k^2q - q + 1 \dots(6)$
Express $q$ in terms: From equation ( $6$ ), we can express $q$ in terms of $z$ and $k$ : $q = \frac{z^2 - 1}{k^2 - 1}$ ...( $7$ ) Since $q$ is a prime number, the numerator must be divisible by the denominator, which means that $z^2 - 1$ must be a multiple of $k^2 - 1$ .
Consider $z + 1$ : The expression $z^2 - 1$ can be factored as $(z + 1)(z - 1)$ . For $q$ to be prime, either $z + 1 = k^2 - 1$ or $z - 1 = k^2 - 1$ , because if both $(z + 1)$ and $(z - 1)$ were multiples of $(k^2 - 1)$ , then $q$ would not be prime as it would have more than two factors.
Substitute $z$ into $q$ : Let's consider the case $z + 1 = k^2 - 1$ : $\newline$ $z + 1 = k^2 - 1$ $\newline$ $z = k^2 - 2$ ...( $8$ )
Consider $z - 1$ : Substitute $z$ from equation ( $8$ ) into equation ( $7$ ): $\newline$ $q = ((k^2 - 2)^2 - 1) / (k^2 - 1)$ $\newline$ This simplifies to: $\newline$ $q = (k^4 - 4k^2 + 4 - 1) / (k^2 - 1)$ $\newline$ $q = (k^4 - 4k^2 + 3) / (k^2 - 1)$ ...( $9$ )
Substitute $z$ into $q$ : For $q$ to be prime, the numerator must be a multiple of the denominator. However, since $k$ is an integer, $k^4 - 4k^2 + 3$ will not be a multiple of $k^2 - 1$ for any integer value of $k$ , except when $k = 1$ , which gives $q = 2$ . But if $k = 1$ , then $q$ $0$ , which contradicts our earlier statement that $q$ $1$ is a multiple of $q$ . Therefore, there are no solutions in this case.
Prime Condition: Now let's consider the case $z - 1 = k^2 - 1$ : $\newline$ $z - 1 = k^2 - 1$ $\newline$ $z = k^2$ ...( $10$ )
Substitute $k$ into $q$ : Substitute $z$ from equation ( $10$ ) into equation ( $7$ ): $\newline$ $q = \frac{(k^{2^2} - 1)}{(k^2 - 1)}$ $\newline$ This simplifies to: $\newline$ $q = \frac{(k^4 - 1)}{(k^2 - 1)}$ $\newline$ $q = k^2 + 1 \quad ...(11)$
Prime Condition: Since $q$ is prime, $k^2 + 1$ must be prime. The only way this can happen is if $k$ is even, because if $k$ is odd, then $k^2 + 1$ is even and greater than $2$ , which cannot be prime. Let's set $k = 2m$ , where $m$ is an integer.
Final Solution: Substitute $k = 2m$ into equation ( $11$ ): $\newline$ $q = (2m)^2 + 1$ $\newline$ $q = 4m^2 + 1$ ...( $12$ )
Final Solution: Substitute $k = 2m$ into equation ( $11$ ):
$q = (2m)^2 + 1$
$q = 4m^2 + 1$ ...( $12$ )For $q$ to be prime, $4m^2 + 1$ must be prime. The only even prime is $2$ , so we must have $m = 0$ to get $q = 1$ , which is not prime, or $m = 1/2$ to get $q = 2$ . However, $q = (2m)^2 + 1$ $0$ must be an integer, so the only solution is $m = 1/2$ , which is not an integer. Therefore, there are no solutions in this case either.