Diophantine equations special problem

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$H(α) = max (|p|, |q|)$.

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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? – Old John Dec 20 '12 at 17:18
@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. – Gerry Myerson Dec 21 '12 at 1:45
@Gerry Myerson! I edited my post in the problem box. Please look.. – gama Dec 21 '12 at 4:26
@Old John! I defined the $H(x)$ at my post. Please look at the main question. – gama Dec 21 '12 at 4:27

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. – Gerry Myerson Dec 22 '12 at 6:21
@Vmrfdu can you give your email ID? – gama Dec 22 '12 at 6:41
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.