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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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    @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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    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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    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