How many solutions?

For how many integer values of x there exists positive integer solutions for SS, such that: S=x(x+p)S=\sqrt{x(x+p)} where x is an integer and p is any fixed prime number >2>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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3 votes

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Here is my solution:

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

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

So, we must have x=N2x=N^2 and x+p=Y2x+p=Y^2 where N,YN, Y are integers.

And now:

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

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

Aditya Parson - 6 years, 3 months ago

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

Siddharth Kumar - 6 years, 3 months ago

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

Aditya Parson - 6 years, 3 months ago

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We have to find the number of integer solutions to the equation x(x+p)=S2 x(x+p) = S^2. First consider the case when x>1 x>1 . Note that gcd(x,x+p)=1 \gcd (x, x+p) = 1 or p 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 \gcd (x, x+p)= p which implies px p \mid x . Let x=pqx= pq , where q q is an integer. Then x(x+p)=pq(p+1)q=q2p(p+1) x(x+p)= pq*(p+1)q = q^2p(p+1) . This has to be a perfect square, and since q2 q^2 is a perfect square, so must be p(p+1) p(p+1) . p(p+1) p(p+1) cannot be a perfect square for p>3 p > 3 . When x=1 x= 1 , p+1 p+1 has to be a perfect square. For all p>3 p>3 , there are no such primes since S21=(S+1)(S1) S^2 - 1= (S+1)(S-1) . Hence (x,p)=(1,3) (x, p)= (1, 3) is the only possible solution.

Sreejato Bhattacharya - 6 years, 3 months ago

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

hero p. - 6 years, 3 months ago

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

Sreejato Bhattacharya - 6 years, 3 months ago

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@Sreejato Bhattacharya The point is that, whenever xx and x+px+p are perfect squares there will exist a solution. It will differ for different values of pp. There will be 22 solutions to the equation for all p>2p> 2.

Aditya Parson - 6 years, 3 months ago

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@Aditya Parson Your phrasing of the original question is vague, because it is not clear whether or not you intend p p to be a fixed prime for all such ordered pairs (x,S) (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={xZ:(p  primeSZ+S=x(x+p))}. 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 x , since for each odd prime p=2m+1 p = 2m+1 for some positive integer m m , there exists at least one integer x=m2 x = m^2 such that S=x(x+p)=m(m+1) S = \sqrt{x(x+p)} = m(m+1) is a positive integer.

hero p. - 6 years, 3 months ago

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@Hero P. Actually, yes. I think I should have stated that pp is any fixed prime >2> 2. Would that do?

Aditya Parson - 6 years, 3 months ago

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

Aditya Parson - 6 years, 3 months ago

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awesome

alpha beta - 6 years, 3 months ago

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i can only prove that S S must be even

  • S2=x(x+p) S^{2}=x(x+p) then S2=x2+xp S^{2}=x^{2}+xp then S2x2=xp S^{2}-x^{2}=xp so S S must be even because if S S is odd,then there is no solution for x x

Tan Li Xuan - 6 years, 3 months ago

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Proving S S is even does not need this much manipulation. All primes greater than 2 2 are odd, so x x and x+p x+p have different parities, thus their product must be even. Hence S2 S^2 is even which implies so is S S .

Sreejato Bhattacharya - 6 years, 3 months ago

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

Aditya Parson - 6 years, 3 months ago

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then whats your method.....?

alpha beta - 6 years, 3 months ago

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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

superman son - 6 years, 3 months ago

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

Aditya Parson - 6 years, 3 months ago

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