Submitted by maths123 on Sat, 05/02/2015 - 17:55

I have used another method to the one listed in the mark scheme for (ii) and (iii) - this method was hinted as being used by candidates in the examiners report. I would be grateful if someone could check this method for (iii) (similar method used in (ii) ):

(iii) From the stem of the question, we know that if there is a rational root then it must be an integer.

Assume there exists an integer $ p $ such that it is a root and hence $ p^n - 5p + 7=0 $

Rearranging we get $ p = \frac {-7}{p^{n-1} - 5} $

since $ p $ is an integer and as 7 is prime, the denominator $ = 1, -1, 7, -7 $ and the corresponding values of $p$, when the values for the denominator are substituted in, are $ -7, 7, -1, 1 $

Thus for the first case:

$ p^{n-1} - 5 = 1 $

$ p^{n-1} = 6 $ - substituting the value $ -7 $ into $p$ we arrive at the contradiction $ (-7)^{n-1} = 6 $ - this is a contradiction since we can take the absolute value of both sides and note $ n-1 \geq 1 $ so $ 7^{n-1} = 6 $ is false

I repeated the above steps for each value of the denominator and corresponding value of $p$ (although I didn't have to take the absolute value to come to the contradiction in the other cases) - since each case leads to a false statement we must conclude that $p$ is not an integer so no rational roots.

## Seems fine from what you've

Seems fine from what you've posted.

The conventional way to use a factorisation argument is not to divide through, but instead argue something like

"we have $p(p^{n-1} - 5) = -7$ and since $p$ is an integer, by prime factorisation, $(p, p^{n-1} - 5)$ must take the values $(\pm 1, \mp 7)$ or $(\pm 7, \mp 1)$."

This is partly because division isn't so nicely behaved (e.g. you need to check for division by zero), and in some more general systems not even defined at all. Though I can see why you wanted to here since you wanted to use directly that $p$ is an integer.