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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?

$(2)$ Is $x(x+d)(x+2d) = y^2$ has infinitely many solutions or not? If there, how to find them.

$(3)$ For $k \ne 2,4$, can we have solutions of $x(x+1)(x+2)...(x+k-1)+Q = t^2$, where Q is a rational number.

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For (1), it is an old theorem of Erdos that in fact the product of $2$ or more consecutive integers can never be a $k$-th power for any $k\ge 2$. –  André Nicolas Dec 24 '12 at 4:43
    
Related: math.stackexchange.com/questions/33338/… –  Zev Chonoles Dec 24 '12 at 4:43
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up vote 2 down vote accepted

One can find infinitely many $d$, $x$, and $y$ such that $x(x+d)(x+2d)=y^2$. Let $w=x+d$. Then we are looking for integers $d$, $w$, and $y$ such that $w(w^2-d^2)=y^2$. Let $(d,t,w)$ be a Pythagorean triple such that $w$ is a perfect square. There are infinitely many such Pythagorean triples.

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Nicolas! what about question (3). –  gama Dec 24 '12 at 5:27
    
Interesting question. There may be some magic identities like the one for $k=4$. If there isn't, the question is undoubtedly difficult. –  André Nicolas Dec 24 '12 at 5:31
    
Nicolas! Please do it for me..dont say difficult. please do it. –  gama Dec 24 '12 at 6:16
    
@I am waiting for (3) question of my post –  gama Dec 24 '12 at 8:34
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It is a type of problem that is notoriously difficult. The result of Erdos was itself not simple to prove, and your question (3) is a large order of magnitude harder, since it mixes multiplication and addition. –  André Nicolas Dec 24 '12 at 8:37
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