Question

Find all rational numbers m for which
\[m+\frac{1}{m}\]
is an integer

Answer 1

\[ = \frac{m^2 + 1}{m}\]

This should be an integer..

Answer 2

-1 and 1 are clear solns..

Answer 3

It will only be integer if \(m^2 + 1\) is entirely divisible by m..

Answer 4

Let m be = z/n (z is an integer and n is a natural number; hint z is not 0)
\[then\ m+1/m = \frac{z}{n}+\frac{n}{z} = \frac{z^2+n^2}{zn}\]

Answer 5

m^2 + 1 = m(m) +1 => divisor(quotient) + rem
only 1 and -1 satisfy the criteria..so i say only 2 solns..

Answer 6

I will go with shubhamsrg..

Answer 7

If m is any number then squaring it and adding it and dividing by the number itself will give you rational values and not integer values..

But for only 1 and -1 it can have Integer results..

Answer 8

thats right but i dont see acceptable reasoninig

Answer 9

See, in \(m + \frac{m}{n}\)

m should be an integer

If m is integer then it reciprocal will not be integer except for values 1 and -1..

Now, sum of two integers will be an integers..

So both m and its reciprocal should be integers and that can only possible if m takes 1 and -1..

Answer 10

Sorry there will come 1/m in place of m/n..

Answer 11

\[\text{for }\frac{z^2+n^2}{zn}\text{ to be an integer, both }z^2 \text{and }n^2\text{must be divisible by n and z respectfully.}\]\[\text{This is of course only possible if z =}\pm n.\text{}\]

Answer 12

how about this
\[m+\frac{1}{m}=t \]
t is an integer
\[m^2+1=mt\\ m^2-mt+1=0 \\ m=\frac{t \pm \sqrt{t^2-4}}{2}\]
inside of radical must be a complete square because if not u will have a Irrational number for m.
t^2-4 is acomplete square if and only if t=2 or t=-2
so m=1 or m=-1

Answer 13

Superb Kazem...