Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.
On
This is a bit cheeky, and probably not what the question meant, but if $m=n$, the condition is satisfied.
On
If I just had to do it for $\frac mn, \frac {m+1}{n+1}, \frac {m+2}{n+2}$ I would note as in lulu's comment that if $m$ and $n$ are both even we satisfy the first and last. If $m$ and $n$ are both one less than a multiple of $3$ we satisfy the second. A couple even numbers that are one less than a multiple of $3$ are $8$ and $14$ so that is one solution. Can you build on this to handle your problem? You will get a system of modular equations and can use the Chinese remainder theorem to make sure there is a solution.
On
Simple computer search gives:
\begin{array}{|c|c|} \hline m & n \\ \hline 212 & 2 \\ 213 & 3 \\ 214 & 4 \\ 215 & 5 \\ 300 & 90 \\ 301 & 91 \\ 324 & 114 \\ 325 & 115 \\ \hline \end{array}
Further observation:
\begin{array}{|c|c|} \hline n \equiv m \pmod{\lambda} & \lambda \\ \hline 2,3,4,5 & 210 \\ 6,7 & 462 \\ 8,9,10,11 & 858 \\ 10,11 & 1430 \\ 12,13 & 1326 \\ 12,13 & 2210 \\ 14,15,16,17 & 1938 \\ 14,15 & 3230 \\ 16,17 & 4522 \\ 18,19 & 2622 \\ 18,19 & 6118 \\ 20,21,22,23 & 690 \\ 20,21 & 1610 \\ 20 & 3795 \\ 24,25 & 870 \\ \hline \end{array}
On
Fixed $n$ a solution is given by
$$m=n+n(n+1)(n+2)(n+3)(n+4)(n+5)=n+\frac{(n+5)!}{(n-1)!}$$
indeed
$$\frac{m}{n}=\frac{n+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n}\\=1+(n+1)(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+1}{n+1}=\frac{n+1+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+1}\\=1+n(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+2}{n+2}=\frac{n+2+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+2}\\=1+n(n+1)(n+3)(n+4)(n+5)$$
$$\frac{m+3}{n+3}=\frac{n+3+n(n+1)(n+3)(n+3)(n+4)(n+5)}{n+3}=\\1+n(n+1)(n+2)(n+4)(n+5)$$
$$\frac{m+4}{n+4}=\frac{n+4+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+4}\\=1+n(n+1)(n+2)(n+3)(n+5)$$
$$\frac{m+5}{n+5}=\frac{n+5+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+5}\\=1+n(n+1)(n+2)(n+3)(n+4)$$
On
I have obtained a very interesting form of fraction which will follow all the conditions. Unfortunately I am not seasoned in number theory and Hence my solution is lengthy, confusing and has a large no. Of variable, and has a few assumptions. It would be helpful if someone could suggest a better and shorter solution. The form is (210k+2)/(420k+2) .
On
I have obtained a very interesting form of fraction which will follow all the conditions. Unfortunately I am not seasoned in number theory and Hence my solution is lengthy, confusing and has a large no. Of variable, and has a few assumptions. It would be helpful if someone could suggest a better and shorter solution. The form is (210k+2)/(420k+2). After a little maths I found that the common factors are the consecutive prime nos. 3,5,7 also when these divide (m+1),(m+3),(m+5) respectively, the remainder is n as m-n is divisible by 3,5,7 .
If $m=30a$ and $n=30b$, then
$${m\over n}={30a\over30b}\\{m+2\over n+2}={2(15a+2)\over2(15b+1)}\\{m+3\over n+3}={3(10a+1)\over3(10b+1)}\\{m+4\over n+4}={2(15a+2)\over2(15b+2)}\\\text{and}\\{m+5\over n+5}={5(6a+1)\over5(6b+1)}$$ are all reducible. It remains to choose $a$ and $b$ so that
$$30a+1\over30b+1$$
is reducible. If we try $b=1$, we see we need $31\mid30a+1$, or $a\equiv1$ mod $31$, so $a=32$ does the trick. Thus $m/n=960/30$ has the desired property.