How many solutions?

For how many integer values of x there exists positive integer solutions for $S$, such that: $S=\sqrt{x(x+p)}$ where x is an integer and p is any fixed prime number $>2$.

A proof will be much better than an integer answer. Since I already know how to solve it, please provide your own proofs. Note by Aditya Parson
6 years, 3 months ago

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Here is my solution:

First let us consider that $x$ and $x+p$ are imperfect squares. Now $gcd (x, x+p)=1$ or $p$. So, if they are imperfect squares there will be no solutions.
Proof:

For $gcd (x, x+p)=p$ we must have $x=ap$ where $a$ is an integer not 0. Doing some manipulation gives us that $S=p\sqrt {a (a+1)}$. Now product of two consecutive integers cannot be a perfect square.

So, we must have $x=N^2$ and $x+p=Y^2$ where $N, Y$ are integers.

And now:

$(Y+N)(Y-N)=p$

Which implies that either of them $=1$ or $p$. Doing some algebra will prove that there are two solutions.

- 6 years, 3 months ago

hint :- square on both sides, do some expansion and then factorize.

- 6 years, 3 months ago

I made the problem, as such I already know the answer.

- 6 years, 3 months ago

We have to find the number of integer solutions to the equation $x(x+p) = S^2$. First consider the case when $x>1$. Note that $\gcd (x, x+p) = 1$ or $p$. But we know that the product of two coprime integers can never be a perfect square, so we must have $\gcd (x, x+p)= p$ which implies $p \mid x$. Let $x= pq$, where $q$ is an integer. Then $x(x+p)= pq*(p+1)q = q^2p(p+1)$. This has to be a perfect square, and since $q^2$ is a perfect square, so must be $p(p+1)$. $p(p+1)$ cannot be a perfect square for $p > 3$. When $x= 1$, $p+1$ has to be a perfect square. For all $p>3$, there are no such primes since $S^2 - 1= (S+1)(S-1)$. Hence $(x, p)= (1, 3)$ is the only possible solution.

- 6 years, 3 months ago

There are several errors in your proof that need correction., since $x = 1, p = 3$ implies $\gcd(1, 4) = 1$, and obviously, $3 \not\mid 1$.

- 6 years, 3 months ago

It has been corrected now. Please check if there are more mistakes remaining.

- 6 years, 3 months ago

The point is that, whenever $x$ and $x+p$ are perfect squares there will exist a solution. It will differ for different values of $p$. There will be $2$ solutions to the equation for all $p> 2$.

- 6 years, 3 months ago

Your phrasing of the original question is vague, because it is not clear whether or not you intend $p$ to be a fixed prime for all such ordered pairs $(x, S)$ satisfying the given relation. As it is written, one could interpret your question so that we are asked to find the size of the set $X = \left\{ x \in \mathbb{Z} : \exists \left( p \; {\rm prime} \cap S \in \mathbb{Z}^+ \cap S = \textstyle\sqrt{x(x+p)} \right) \right\}.$ In this case, there are infinitely many such $x$, since for each odd prime $p = 2m+1$ for some positive integer $m$, there exists at least one integer $x = m^2$ such that $S = \sqrt{x(x+p)} = m(m+1)$ is a positive integer.

- 6 years, 3 months ago

Actually, yes. I think I should have stated that $p$ is any fixed prime $> 2$. Would that do?

- 6 years, 3 months ago

After stating that $gcd (x, x+p)=1$ or $p$, you should have considered that $S$ will have a solution when $x$ and $x+p$ are both perfect squares.

- 6 years, 3 months ago

awesome

- 6 years, 3 months ago

i can only prove that $S$ must be even

• $S^{2}=x(x+p)$ then $S^{2}=x^{2}+xp$then $S^{2}-x^{2}=xp$ so $S$ must be even because if $S$ is odd,then there is no solution for $x$

- 6 years, 3 months ago

Proving $S$ is even does not need this much manipulation. All primes greater than $2$ are odd, so $x$ and $x+p$ have different parities, thus their product must be even. Hence $S^2$ is even which implies so is $S$.

- 6 years, 3 months ago

Your statement that $S$ must be even is correct, but I did not approach the problem this way. .

- 6 years, 3 months ago

then whats your method.....?

- 6 years, 3 months ago

i think for every prime there is a solution of x eg- (1,3)(4,5)(9,7)(11,25) i.e x= (p^2 -2*p +1)/4

- 6 years, 3 months ago

The prime is fixed, and for every odd prime there are 2 solutions.

- 6 years, 3 months ago