solutions of consecutive integers and observations

I wonder that, I made these observations from my previous study on product of consecutive integers. I am looking the solutions of these kind of equations.

$(1)$ Is $x(x+1)(x+2)...(x+[\text{any-odd-integer}]) = y^2$ has solutions or not?. If exits, how to list them?

$(2)$Prove that For any $k \ne 2,4$,a polynomial function of a form $x(x+1)(x+2)...(x+k-1)+Q = t^2$, where Q is a rational number has solutions or not? here $x$ is variable and others maybe constants. Your answer is very essential for me.

-

Regarding you first question: no this does not have any solution by a result of Erdős "Note on product of consecutive integers" (Journal London Math. Soc. 1939); you do not need the condition 'odd' either.

I do not understand the second question. For some $Q$ of course there will be some solution, for others not; but in general it is unclear to me what quantities are the variables and which are to be considered constant.

Added (in view of the other answer/comment): There are various investigations around this; a somewhat recent paper that seems still closer to this question would be Bilu, Kulkarni, Sury "The Diophantine equation $x(x+1) \dots (x+ (m-1))+ r=y^n$" (note the $n$ can be $2$).

A result from there implies that (coming back to the question, so $n=2$) if $r$ is not a square, then there are only finitely many solutions (and these can be determined explicitly). And other results are mentioned. In general, there is quite a bit of literature on this type of question. Without further information from OP it is not really clear what type of information would be relevant.

-
@Quid! Here Q is any rational number. – gama Dec 27 '12 at 6:16
Yes, I understood that much. But in your equation what is to be considered as a constant and what as a variable. Or, what type of information regarding 'solutions' are you looking for precisely. As I said, for some rational Q of course there is a solution and for others not. So, what do you want to know precisely? – quid Dec 27 '12 at 8:13
@Qquid! Prove that $For any$k \ne 2,4$,a polynomial function of a form$x(x+1)(x+2)...(x+k-1)+Q = t^2$, where Q is a rational number has solutions or not? here$x$is variable and others maybe constants. Your answer is very essential for me. – gama Dec 27 '12 at 9:18 @Quid! I edited my main question at main post. please see – gama Dec 27 '12 at 9:46 Please note that this question has already been posted here on Math.stackexchange. You seem to be trying to extend the result from this paper: "On Diophantine equations of the form$(x−a1)(x−a_2)…(x−a_k)+r=y^n.$" by Manisha Kulkarni and B.Sury, which is available here to the case$n=2\$.

(This should probably have been a comment, but I don't think I have enough reputation)

-
Thanks! To know this is quite useful – quid Dec 24 '12 at 12:55
Highly disreputable. – Will Jagy Dec 24 '12 at 18:23
@Will Jagy: in gama's defence, they waited some hours and got a partal response and the info it was difficult. And the current question dropped the part which got answered. It is certainly not optimal how the thing got 'crossposted', but quite a bit better than we often see it. – quid Dec 24 '12 at 20:48