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Suppose $r$ is a rational number and for $k > 2$, consider $0\leqslant a_1< a_2<\cdots \leqslant a_k$. Also, for $n > 2$ and assume that we are not interesting the case of $n = 4 = k$, then there exists only finitely many solutions of $x$ in set of integers and $y$ in set of rational numbers to the equation $$ r + (x-a_1)(x-a_2)\cdots(x-a_k) = y^n $$ and all the solutions satisfy $\max\{H(x), H(y)\} < C$, where $C$, is an effectively computable constant depending only on $n$, $r$, and $a_i$'s. Here $r$ is an integer and not a perfect $n$-th power. Generalize the truth of this statement and show the solutions existence with $k$ bound.

$edit$: We recall that the height $H(α)$ of an algebraic number α is the maximum of the absolute values of the integer coefficients in its minimal defining polynomial In particular, if α is a rational integer, then $H(α) = |α|$ and if α is a rational number and not equal to $zero$ Then$ H(α) = max (|p|, |q|)$.

Advanced thanks and Happy Christmas...

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    Can you explain what you mean by $H(x)$? Can you also give some information about what progress you have made with this, where you are stuck, and where this problem comes from?2012-12-20
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    @Old, probably $H$ is the *height*. For an integer $x$, it's just $|x|$; for a rational $p/q$, it's $\max(|p|,|q|)$. But, yes, it would be nice if OP would be a bit more forthcoming here.2012-12-21
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    @Gerry Myerson! I edited my post in the problem box. Please look..2012-12-21
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    @Old John! I defined the $H(x)$ at my post. Please look at the main question.2012-12-21

2 Answers 2

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Let us denote Pk,c(x) = x(x + 1)(x + 2) . . . (x + k − 1) + c. Suppose that Pk,c(x) = a(x)^2, k = 2n. Then

Pk,c(x + 1) − Pk,c(x) = k(x + 1)(x + 2) . . . (x + k − 1) = a(x + 1)^2 − a(x)^2 .

Implies that (a(x + 1) − a(x))(a(x + 1) + a(x)) = k(x + 1)(x + 2) . . . (x + k − 1) .

As the graph of the polynomial y = a(x+1) could be obtained by translation to the left by 1 from the graph y = a(x), each of n − 1 solutions of the equation a(x + 1) = a(x) lies between a pair of roots of the polynomial a(x) + a(x + 1) (which have n roots).

Hence a(x + 1) − a(x) = n(x + 2)(x + 4) . . . (x + 2n − 2) , a(x + 1) + a(x) = 2(x + 1)(x + 3) . . . (x + 2n − 1) .

By addition, we get; 2a(x + 1) = 2(x + 1)(x + 3) . . . (x + 2n − 1) + n(x + 2)(x + 4) . . . (x + 2n − 2) . And substituting the same changing x by x + 1 we obtain 2a(x + 1) = 2(x + 2)(x + 4) . . . (x + 2n) − n(x + 3)(x + 5) . . . (x + 2n − 1) . Two obtained expressions contradict to each other. To be ensure this put x = 0 to both and subtract one from another. We get ;

(n + 2)(1 · 3 · · · (2n − 1)) = 3n(2 · 4 · · · (2n − 2)),

Here the right hand contains two as a factor with more power than left hand side.

If you can refer Tchebyshev Theorem and Bertrand Postulate, you will get complete data and you can understand well about my script. In case of further assistance, you can write your comments.

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    The formula for $P(x+1)-P(x)$ is wrong. Even if it were right, the formulas for $a(x+1)\pm a(x)$ don't follow from it.2012-12-22
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    @Vmrfdu can you give your email ID?2012-12-22
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    GAMA..What your you want ask, ask here itself. Others may answer.2012-12-22
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    @GerryMyerson! you are correct..let me answer once again.2012-12-22
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    GAMA! ask you question here itself...don't ask personal email ID. Not only me many will answer, once you post your question here itself.2012-12-22
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    GAMA! I am little poor in LATEX/TEX typing. I hope you can understand!2012-12-22
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    @vmrfdu123456! can you give please your email ID2012-12-22
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    GAMA! As i told you, I am not intended to write my email ID here. Thank you so much for accepting my answer. However, you need to understand more...not only my solution.2012-12-22
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    You seem to be assuming that $P_{k,c}$ has $2n-1$ roots. Why? It might not have any real roots at all, if $c$ is big.2012-12-23
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This seems to be an exact word-for-word copy of Theorem 1 from the paper "On Diophantine equations of the form $(x − a_1)(x − a_2) \dots (x − a_k) + r = y^n$" by Manisha Kulkarni and B.Sury.

The paper containing the proof is available here: http://www.isibang.ac.in/~statmath/eprints/2011/6.pdf.

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    ! The proof is given at the paper is quite uninteresting2012-12-21
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    @gama Could you explain exactly why you find the proof given in the paper to be quite uninteresting? It seems to be quite clear and concise.2012-12-21
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    unnecessary discussions are given in theorem 2 and so on. I am searching more precise solution rather than the paper, which you have given.2012-12-21
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    It is difficult to me to explain the uninteresting factor by typing. I am sure, becoz of your paper many will read the question and they express the same, what I felt2012-12-21
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    I would suggest that if you are looking for improvements for published research-level results such as the above, then you might be more likely to get better results from asking your question at [mathoverflow.net](http://mathoverflow.net/).2012-12-21
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    I would suggest writing to an author of the paper. The authors would be the experts on the topic, and in the best position to answer your questions. (Just don't tell them that you found their proof uninteresting!)2012-12-21