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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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.