Mixing
For integers $a,b$ and prime $p$, are there infinitely many solutions to $p = (a-b)(a+b)$?
$begingroup$
I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p mid (a-b)$ or $p mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:
For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
$$p = (a-b)(a+b)$$
number-theory elementary-number-theory prime-numbers
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
$endgroup$
add a comment |
$begingroup$
I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p mid (a-b)$ or $p mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:
For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
$$p = (a-b)(a+b)$$
number-theory elementary-number-theory prime-numbers
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
$endgroup$
$begingroup$
You have written down all the equations needed for the solution! Just list some of the solutions, and figure out the pattern.
$endgroup$
– Trebor
Jan 6 at 3:33
1
$begingroup$
You have it backwards. $p = (a-b)(a+b) $ does *NOT* mean $p|a+b$ and $p|a-b$. It means $a+b|p$ and $a-b|p$. Which means $a+b = pm 1, pm p$ and $a-b = pm p, pm 1$. There are only four solutions. If we assume $a$ and $b$ are positive there is only one.
$endgroup$
– fleablood
Jan 6 at 4:30
1
$begingroup$
For every odd $p$ you have a solution $a+b = p$ and $a-b = 1$. That just means $a = frac {p+1}2$ and $b = frac {p-1}2$. This is, of course true for all odd $p$. For every $p$ there is exactly one such positive solution.
$endgroup$
– fleablood
Jan 6 at 4:33
add a comment |
$begingroup$
I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p mid (a-b)$ or $p mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:
For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
$$p = (a-b)(a+b)$$
number-theory elementary-number-theory prime-numbers
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
$endgroup$
I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p mid (a-b)$ or $p mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:
For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
$$p = (a-b)(a+b)$$
number-theory elementary-number-theory prime-numbers
number-theory elementary-number-theory prime-numbers
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
edited Jan 6 at 3:48
Eevee Trainer
5,7871936
5,7871936
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
asked Jan 6 at 3:28
Richard KingstonRichard Kingston
233
233
$begingroup$
You have written down all the equations needed for the solution! Just list some of the solutions, and figure out the pattern.
$endgroup$
– Trebor
Jan 6 at 3:33
1
$begingroup$
You have it backwards. $p = (a-b)(a+b) $ does *NOT* mean $p|a+b$ and $p|a-b$. It means $a+b|p$ and $a-b|p$. Which means $a+b = pm 1, pm p$ and $a-b = pm p, pm 1$. There are only four solutions. If we assume $a$ and $b$ are positive there is only one.
$endgroup$
– fleablood
Jan 6 at 4:30
1
$begingroup$
For every odd $p$ you have a solution $a+b = p$ and $a-b = 1$. That just means $a = frac {p+1}2$ and $b = frac {p-1}2$. This is, of course true for all odd $p$. For every $p$ there is exactly one such positive solution.
$endgroup$
– fleablood
Jan 6 at 4:33
add a comment |
$begingroup$
You have written down all the equations needed for the solution! Just list some of the solutions, and figure out the pattern.
$endgroup$
– Trebor
Jan 6 at 3:33
1
$begingroup$
You have it backwards. $p = (a-b)(a+b) $ does *NOT* mean $p|a+b$ and $p|a-b$. It means $a+b|p$ and $a-b|p$. Which means $a+b = pm 1, pm p$ and $a-b = pm p, pm 1$. There are only four solutions. If we assume $a$ and $b$ are positive there is only one.
$endgroup$
– fleablood
Jan 6 at 4:30
1
$begingroup$
For every odd $p$ you have a solution $a+b = p$ and $a-b = 1$. That just means $a = frac {p+1}2$ and $b = frac {p-1}2$. This is, of course true for all odd $p$. For every $p$ there is exactly one such positive solution.
$endgroup$
– fleablood
Jan 6 at 4:33
$begingroup$
You have written down all the equations needed for the solution! Just list some of the solutions, and figure out the pattern.
$endgroup$
– Trebor
Jan 6 at 3:33
$begingroup$
You have written down all the equations needed for the solution! Just list some of the solutions, and figure out the pattern.
$endgroup$
– Trebor
Jan 6 at 3:33
1
1
$begingroup$
You have it backwards. $p = (a-b)(a+b) $ does *NOT* mean $p|a+b$ and $p|a-b$. It means $a+b|p$ and $a-b|p$. Which means $a+b = pm 1, pm p$ and $a-b = pm p, pm 1$. There are only four solutions. If we assume $a$ and $b$ are positive there is only one.
$endgroup$
– fleablood
Jan 6 at 4:30
$begingroup$
You have it backwards. $p = (a-b)(a+b) $ does *NOT* mean $p|a+b$ and $p|a-b$. It means $a+b|p$ and $a-b|p$. Which means $a+b = pm 1, pm p$ and $a-b = pm p, pm 1$. There are only four solutions. If we assume $a$ and $b$ are positive there is only one.
$endgroup$
– fleablood
Jan 6 at 4:30
1
1
$begingroup$
For every odd $p$ you have a solution $a+b = p$ and $a-b = 1$. That just means $a = frac {p+1}2$ and $b = frac {p-1}2$. This is, of course true for all odd $p$. For every $p$ there is exactly one such positive solution.
$endgroup$
– fleablood
Jan 6 at 4:33
$begingroup$
For every odd $p$ you have a solution $a+b = p$ and $a-b = 1$. That just means $a = frac {p+1}2$ and $b = frac {p-1}2$. This is, of course true for all odd $p$. For every $p$ there is exactly one such positive solution.
$endgroup$
– fleablood
Jan 6 at 4:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose there exists integers $a,b$ and prime $p$ such that
$$p = (a-b)(a+b)$$
This implies that $(a-b)$ and $(a+b)$ are factors of $p$. Since $p$ is prime, then one of $(a-b)$ and one $(a+b)$ must be $1$ and the other $p$, otherwise $p$ is not prime. Thus, one of the two systems must be the case:
$$left{begin{matrix}
a+b = 1 \
a-b = p
end{matrix}right. ;;; text{or, alternatively} ;;; left{begin{matrix}
a-b = 1 \
a+b = p
end{matrix}right.$$
Solve the system on the left: you'll find $a = (p+1)/2, b = (1-p)/2$.
Solve the system on the right: you'll find $a = (p+1)/2, b = (p-1)/2$.
Thus, for any given $p$, there are only two solutions for $a,b$ in integers. provided $p>2$. If $p=2$, then neither $a$ or $b$ are integers.
If you mean solutions in terms of all three variables - $a,b,p$ - then there are infinitely many since you can relate $a,b$ in terms of any given $p>2$.
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=frac{p+1}{2}$ for every prime p besides $2$.
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
$endgroup$
add a comment |
$begingroup$
Of course not!
If $p$ is prime so its only factors are $pm 1$ and $pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $pm 1$ and/or $pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=frac {p+1}2; b = frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= frac {p+1}2; b = -frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-frac {p+1}2; b=-frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-frac{p+1}2; b = frac {p-1}2 = -1-a$
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
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$begingroup$
Suppose there exists integers $a,b$ and prime $p$ such that
$$p = (a-b)(a+b)$$
This implies that $(a-b)$ and $(a+b)$ are factors of $p$. Since $p$ is prime, then one of $(a-b)$ and one $(a+b)$ must be $1$ and the other $p$, otherwise $p$ is not prime. Thus, one of the two systems must be the case:
$$left{begin{matrix}
a+b = 1 \
a-b = p
end{matrix}right. ;;; text{or, alternatively} ;;; left{begin{matrix}
a-b = 1 \
a+b = p
end{matrix}right.$$
Solve the system on the left: you'll find $a = (p+1)/2, b = (1-p)/2$.
Solve the system on the right: you'll find $a = (p+1)/2, b = (p-1)/2$.
Thus, for any given $p$, there are only two solutions for $a,b$ in integers. provided $p>2$. If $p=2$, then neither $a$ or $b$ are integers.
If you mean solutions in terms of all three variables - $a,b,p$ - then there are infinitely many since you can relate $a,b$ in terms of any given $p>2$.
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
Suppose there exists integers $a,b$ and prime $p$ such that
$$p = (a-b)(a+b)$$
This implies that $(a-b)$ and $(a+b)$ are factors of $p$. Since $p$ is prime, then one of $(a-b)$ and one $(a+b)$ must be $1$ and the other $p$, otherwise $p$ is not prime. Thus, one of the two systems must be the case:
$$left{begin{matrix}
a+b = 1 \
a-b = p
end{matrix}right. ;;; text{or, alternatively} ;;; left{begin{matrix}
a-b = 1 \
a+b = p
end{matrix}right.$$
Solve the system on the left: you'll find $a = (p+1)/2, b = (1-p)/2$.
Solve the system on the right: you'll find $a = (p+1)/2, b = (p-1)/2$.
Thus, for any given $p$, there are only two solutions for $a,b$ in integers. provided $p>2$. If $p=2$, then neither $a$ or $b$ are integers.
If you mean solutions in terms of all three variables - $a,b,p$ - then there are infinitely many since you can relate $a,b$ in terms of any given $p>2$.
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
Suppose there exists integers $a,b$ and prime $p$ such that
$$p = (a-b)(a+b)$$
This implies that $(a-b)$ and $(a+b)$ are factors of $p$. Since $p$ is prime, then one of $(a-b)$ and one $(a+b)$ must be $1$ and the other $p$, otherwise $p$ is not prime. Thus, one of the two systems must be the case:
$$left{begin{matrix}
a+b = 1 \
a-b = p
end{matrix}right. ;;; text{or, alternatively} ;;; left{begin{matrix}
a-b = 1 \
a+b = p
end{matrix}right.$$
Solve the system on the left: you'll find $a = (p+1)/2, b = (1-p)/2$.
Solve the system on the right: you'll find $a = (p+1)/2, b = (p-1)/2$.
Thus, for any given $p$, there are only two solutions for $a,b$ in integers. provided $p>2$. If $p=2$, then neither $a$ or $b$ are integers.
If you mean solutions in terms of all three variables - $a,b,p$ - then there are infinitely many since you can relate $a,b$ in terms of any given $p>2$.
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
Suppose there exists integers $a,b$ and prime $p$ such that
$$p = (a-b)(a+b)$$
This implies that $(a-b)$ and $(a+b)$ are factors of $p$. Since $p$ is prime, then one of $(a-b)$ and one $(a+b)$ must be $1$ and the other $p$, otherwise $p$ is not prime. Thus, one of the two systems must be the case:
$$left{begin{matrix}
a+b = 1 \
a-b = p
end{matrix}right. ;;; text{or, alternatively} ;;; left{begin{matrix}
a-b = 1 \
a+b = p
end{matrix}right.$$
Solve the system on the left: you'll find $a = (p+1)/2, b = (1-p)/2$.
Solve the system on the right: you'll find $a = (p+1)/2, b = (p-1)/2$.
Thus, for any given $p$, there are only two solutions for $a,b$ in integers. provided $p>2$. If $p=2$, then neither $a$ or $b$ are integers.
If you mean solutions in terms of all three variables - $a,b,p$ - then there are infinitely many since you can relate $a,b$ in terms of any given $p>2$.
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
answered Jan 6 at 3:41
Eevee TrainerEevee Trainer
5,7871936
5,7871936
add a comment |
add a comment |
$begingroup$
If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=frac{p+1}{2}$ for every prime p besides $2$.
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
$endgroup$
add a comment |
$begingroup$
If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=frac{p+1}{2}$ for every prime p besides $2$.
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
$endgroup$
add a comment |
$begingroup$
If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=frac{p+1}{2}$ for every prime p besides $2$.
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
$endgroup$
If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=frac{p+1}{2}$ for every prime p besides $2$.
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
answered Jan 6 at 3:37
Erik ParkinsonErik Parkinson
8549
8549
add a comment |
add a comment |
$begingroup$
Of course not!
If $p$ is prime so its only factors are $pm 1$ and $pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $pm 1$ and/or $pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=frac {p+1}2; b = frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= frac {p+1}2; b = -frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-frac {p+1}2; b=-frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-frac{p+1}2; b = frac {p-1}2 = -1-a$
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
$begingroup$
Of course not!
If $p$ is prime so its only factors are $pm 1$ and $pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $pm 1$ and/or $pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=frac {p+1}2; b = frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= frac {p+1}2; b = -frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-frac {p+1}2; b=-frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-frac{p+1}2; b = frac {p-1}2 = -1-a$
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
$begingroup$
Of course not!
If $p$ is prime so its only factors are $pm 1$ and $pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $pm 1$ and/or $pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=frac {p+1}2; b = frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= frac {p+1}2; b = -frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-frac {p+1}2; b=-frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-frac{p+1}2; b = frac {p-1}2 = -1-a$
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
$endgroup$
Of course not!
If $p$ is prime so its only factors are $pm 1$ and $pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $pm 1$ and/or $pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=frac {p+1}2; b = frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= frac {p+1}2; b = -frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-frac {p+1}2; b=-frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-frac{p+1}2; b = frac {p-1}2 = -1-a$
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
answered Jan 6 at 4:27
fleabloodfleablood
69.3k22685
69.3k22685
add a comment |
add a comment |
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$begingroup$
You have written down all the equations needed for the solution! Just list some of the solutions, and figure out the pattern.
$endgroup$
– Trebor
Jan 6 at 3:33
1
$begingroup$
You have it backwards. $p = (a-b)(a+b) $ does *NOT* mean $p|a+b$ and $p|a-b$. It means $a+b|p$ and $a-b|p$. Which means $a+b = pm 1, pm p$ and $a-b = pm p, pm 1$. There are only four solutions. If we assume $a$ and $b$ are positive there is only one.
$endgroup$
– fleablood
Jan 6 at 4:30
1
$begingroup$
For every odd $p$ you have a solution $a+b = p$ and $a-b = 1$. That just means $a = frac {p+1}2$ and $b = frac {p-1}2$. This is, of course true for all odd $p$. For every $p$ there is exactly one such positive solution.
$endgroup$
– fleablood
Jan 6 at 4:33