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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
4 years, 1 month 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. Aditya Parson · 4 years, 1 month ago

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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. Sreejato Bhattacharya · 4 years, 1 month ago

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@Sreejato Bhattacharya 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. Aditya Parson · 4 years, 1 month ago

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@Sreejato Bhattacharya There are several errors in your proof that need correction., since \( x = 1, p = 3 \) implies \( \gcd(1, 4) = 1 \), and obviously, \( 3 \not\mid 1 \). Hero P. · 4 years, 1 month ago

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@Hero P. It has been corrected now. Please check if there are more mistakes remaining. Sreejato Bhattacharya · 4 years, 1 month ago

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@Sreejato Bhattacharya 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 \(p> 2\). Aditya Parson · 4 years, 1 month 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 \( p \) 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. Hero P. · 4 years, 1 month ago

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@Hero P. Actually, yes. I think I should have stated that \(p\) is any fixed prime \(> 2\). Would that do? Aditya Parson · 4 years, 1 month ago

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@Sreejato Bhattacharya awesome Alpha Beta · 4 years, 1 month ago

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hint :- square on both sides, do some expansion and then factorize. Siddharth Kumar · 4 years, 1 month ago

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@Siddharth Kumar I made the problem, as such I already know the answer. Aditya Parson · 4 years, 1 month 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 · 4 years, 1 month ago

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@Superman Son The prime is fixed, and for every odd prime there are 2 solutions. Aditya Parson · 4 years, 1 month ago

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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 \( x \)
Tan Li Xuan · 4 years, 1 month ago

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@Tan Li Xuan 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 \( S \). Sreejato Bhattacharya · 4 years, 1 month ago

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@Tan Li Xuan Your statement that \(S\) must be even is correct, but I did not approach the problem this way. . Aditya Parson · 4 years, 1 month ago

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@Aditya Parson then whats your method.....? Alpha Beta · 4 years, 1 month ago

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