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How to prove that, there does not exist any positive integers pair $(m,n)$ satisfying:

$n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$.

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    it is clear that,if this function have a solution,then $n>m$ or $n=m*t+q$ but put this value makes much complex this equation2012-07-10
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    if you put or expand powers,this may help http://www.wolframalpha.com/input/?i=n*%28n%2B1%29*%28n%2B2%29*%28n%2B3%29%3Dm*%28m%2B1%29%5E2*%28m%2B2%29%5E3*%28m%2B4%29%5E4++n%3E0%2Cm%3E02012-07-10
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    Where does the problem come from? What do you know about it?2012-07-10
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    Someone give it to me ,i was curious if there is some specific method to solve like this problem.2012-07-10
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    Given the number of +1 to @GerryMyerson's comment I guess that I am not the only person to realise that there is no totally standard/easy way to do it. Depending on where this question comes from, it might be worth continueing to search for an elementary solution or to use heavier machinery.2012-07-10
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    Clearly, the R.H.S. ≥ $1.2^2.3^3.4^4$=27648, $27648≤ n(n+1)(n+2)(n+3)≤ (n+3)^4$ n+3≥12.89... But if n=11, the L.H.S. = 11*12*13*14 = 24 024 =>n≥12.2012-07-10
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    For any prime factor(p>3) of m+3, i)$p^4$ must divide exactly one of the factors of L.H.S. as the G.C.D. of any of the two factors in the L.H.S. ≤3 < p. Similarly, (ii)for any prime factor(q≥3) of m+2, $q^3$ must divide exactly one of the factors of L.H.S. and (iii)for any prime factor(r≥3) of m+1, $r^2$ must divide exactly one of the factors of L.H.S. as (m+1) or (m+2) can have GCD ≤2 with the rest factors of L.H.S.2012-07-10
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    Also observe, n(n+1)(n+2)(n+3) = $(n^2+3n-1)^2-1$2012-07-10

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