# Divisibility, many unknowns

As I was solving a math problem, I stumbled upon a question:

What conditions must $a,b,c,d,e$ meet for $n$ to be a natural number?

(Frankly speaking, I would like all but one of $a,b,c,d,e$ to be equal to $1$, but I do not know how to prove it.)

Here it is:

$$n=\frac{bcde+cde+de+e+1}{abcde-1}$$

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You cannot have everything equal to 1, because everything equal to 2 is a solution. – Phira Oct 31 '11 at 10:14
I am sorry for this imprecision. The trick is , n>1. – MickeyMouse Oct 31 '11 at 10:44
If this is the "trick", could you put this "trick" in the formulation of the question? – Phira Oct 31 '11 at 10:52

Note that if we denote your expression $f(a,b,c,d,e)$ then $$an=\frac{abcde+acde+ade+ae+a}{abcde-1}= 1+\frac{acde+ade+ae+a+1}{abcde-1} =1+f(b,c,d,e,a)$$

When $a=1$ the expression decreases by 1, when $a$ is at least 2, the expression increases at least by $a-2$. The same argument holds for the other letters.

If one of the letters is 2, the expression is reduced to only four variables (note that the integrality of all cyclic shifts is equivalent).

So, now you can try $(6,1,1,1,1)$, $(5,1,1,1,1)$, $(4,1,1,1,1)$, $(3,1,1,1,1)$, $(3,3,1,1,1)$, $(3,1,3,1,1)$, $(5,1,1,1)$,$(4,1,1,1)$,$(3,1,1,1)$,$(3,3,1,1)$,$(3,1,3,1)$, $(4,1,1)$,$(3,1,1)$,$(3,1)$.

For every solution you find, you can insert an appropriate number of twos, including the solution with only twos.

For example, you check that $(3,1)$ has indeed the property that $\frac{3+1}{1\cdot 3-1}$ is integral, therefore $(3,1,2,2,2)$, $(1,2,2,2,3)$, $(3,2,1,2,2)$, etc are solutions of the original equation.

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Thank you. It is really helpful. However, I don't know what (5,1,1,1),(4,1,1,1)...(3,1,1),(3,1) mean. There are 5 variables, aren't there? – MickeyMouse Oct 31 '11 at 11:06
"you can insert an appropriate number of twos" – Gerry Myerson Oct 31 '11 at 11:27
Ah, right, indeed. – MickeyMouse Oct 31 '11 at 11:36