# Prove that there does not exist any positive integers pair $(m, n)$ satisfying: $n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$

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

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

• 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 equation Jul 10, 2012 at 12:21
• if you put or expand powers,this may help 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%3E0 Jul 10, 2012 at 12:29
• Where does the problem come from? What do you know about it? Jul 10, 2012 at 13:13
• Someone give it to me ,i was curious if there is some specific method to solve like this problem. Jul 10, 2012 at 15:18
• 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. Jul 10, 2012 at 15:34

This is an edited version of a partial answer that I posted sometime ago and subsequently deleted (not sure if resurrecting an answer is the correct thing to do after it has been up-voted and then deleted, perhaps someone will advise). If anyone can suggest where any of this can be improved, or point out any mistakes, I would be grateful.

Consider the equation $$n(n+1)(n+2)(n+3) = m(m+1)^2(m+2)^3(m+3)^4$$ To avoid some trivialities later on, it is easy to check that there are no solutions with $m=1$ or $m=2$.

Using the fact that $n(n+1)(n+2)(n+3)$ is almost a square, we have $$(n^2+3n+1)^2-1 = m(m+2)\times[(m+1)(m+2)(m+3)^2]^2$$ Putting $N = n^2+3n+1$ and $M = (m+1)(m+2)(m+3)^2$, this becomes $$N^2-1 = m(m+2)M^2$$ so that $$N^2-1 = [(m+1)^2-1]M^2.$$ Our approach now is to write this as $$N^2 - [(m+1)^2-1]M^2 = 1,$$ which is Pell's equation, with $d = (m+1)^2-1$. In this case there is a particularly nice solution for the Pell equation, as the continued fraction is very simple in this instance. For convenience we change notation slightly and use $k = m+1$, so that we are looking at solutions of $$x^2 - (k^2-1)y^2 = 1,$$ and bear in mind that for any solution $(x,y)$ we also require $$y = (m+1)(m+2)(m+3)^2 = k(k+1)(k+2)^2.$$ So we investigate the properties of solutions of the Pell equation above by looking at the standard continued fraction method. We have $$\sqrt{k^2-1} = (k-1)+\cfrac{1}{1+\cfrac{1}{(2k-2)+\cfrac{1}{1+\cfrac{1}{(2k-2) + \ddots}}}}$$ which gives the first few solutions $(x_n,y_n)$ as $(1,0), (k,1), (2k^2-1,2k), \dots$.

Looking at the solutions for $y$, we see that they are generated by the recurrence relation $$y_{n+2} = 2ky_{n+1} - y_n, \mbox{ with } y_0 = 0, y_1 = 1.$$ Recalling that we also need $y = k(k+1)(k+2)^2$, it is enough to prove that this last expression cannot be one of the $y_n$ from the recurrence relation as follows. Clearly, the values $y_n$ are strictly increasing, and we claim that $$y_4 < k(k+1)(k+2)^2 < y_5$$ A bit of algebra gives $$y_4 = 8k^3-4k$$ so that \begin{equation*}k(k+1)(k+2)^2-y_4 = k^4-3k^3+8k^2+8k = k^3(k-3)+8k^2+8k\end{equation*} which is clearly positive for $k\geq 3$ and is easily checked to be positive for $k=1,2$.

For the right-hand inequality above, we have $$y_5 = 16k^4-12k^2+1$$ and then $$y_5 - k(k+1)(k+2)^2 = 16k^4-12k^2+1 - (k^4+5k^3+8k^2+4k)$$ $$= 15k^4-5k^3-20k^2-4k+1$$ and by examining the graph (because I can't see an elegant way to do this bit) we see that this is positive for $k\geq 2$, and we know that $k=1$ (corresponding to $m=2$) is not a solution of the original equation.

This shows that $k(k+1)(k+2)^2$ cannot be one of the $y_n$, so that no solution of the original equation is possible.

I am sure that there ought to be a simpler solution, but I have been unable to find one.