# 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.

5 years, 9 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.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

- 5 years, 9 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.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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

- 5 years, 9 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 $$p> 2$$.

- 5 years, 9 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.

- 5 years, 9 months ago

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

- 5 years, 9 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.

- 5 years, 9 months ago

awesome

- 5 years, 9 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 $$x$$

- 5 years, 9 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 $$S$$.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

- 5 years, 9 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

- 5 years, 9 months ago

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

- 5 years, 9 months ago