Interesting proof question.

Note that:
$\sqrt 2 - 1 = \sqrt 2 - \sqrt 1$
$(\sqrt 2 - 1)^2 = \sqrt 9 - \sqrt 8$
$(\sqrt 2 - 1 )^3 = \sqrt{50} - \sqrt{49}$
$(\sqrt 2 - 1 )^4 = \sqrt{289} - \sqrt{288}$
And so on.
So we can hypothesise that, for any positive integer $n$, $(\sqrt 2 - 1 )^n$ can be expressed as the difference of the square roots of two consecutive integers.
How would we prove this?
I've tried things but I can't seem to get anywhere. I don't think I can use induction for this because the consecutive rooted integers seem to have no pattern between them as n increases (at least I haven't spotted it yet). Then again, it is fairly late so I'm not at my sharpest.
Thanks in advance.

Binomial expansion, or proof by induction incorporating a binomial expansion? Something like that might be worth a bash. That way you get to use (Rt2 - 1)^n using your proposition, then multiply it by (Rt2 - 1). See what happens. (I'm doubting that a direct binomial expansion will have something nice pop out ... but you never know!)

Not sure this gets me anywhere of interest when trying to show that it should give a difference of two rooted consecutive integers though since those to integers could be anything (AFAIK anyways).

notice that 1 of the numbers in the root is always a square number of a prime ie. 17x17=289 etc.. excluding two or they all are square numbers including 1. now prove it

Note that:
$\sqrt 2 - 1 = \sqrt 2 - \sqrt 1$
$(\sqrt 2 - 1)^2 = \sqrt 9 - \sqrt 8$
$(\sqrt 2 - 1 )^3 = \sqrt{50} - \sqrt{49}$
$(\sqrt 2 - 1 )^4 = \sqrt{289} - \sqrt{288}$
And so on.
So we can hypothesise that, for any positive integer $n$, $(\sqrt 2 - 1 )^n$ can be expressed as the difference of the square roots of two consecutive integers.
How would we prove this?
I've tried things but I can't seem to get anywhere. I don't think I can use induction for this because the consecutive rooted integers seem to have no pattern between them as n increases (at least I haven't spotted it yet). Then again, it is fairly late so I'm not at my sharpest.
Thanks in advance.

(1-sqrt(2))^n can be written as x + y sqrt(2) (*) with x y integers.
But then (1+sqrt(2)) = x - y sqrt 2. Multiply the two together and we have (-1)^n = x^2 - 2y^2. Since we can rewrite x + y sqrt(2) as sqrt(x^2) + sqrt(2y^2) the result follows.

Note also that by considering what happens when you multiply (*) by (1 - sqrt 2) you can get a Fibonnaci like relationship between successive values of x and y. (so you can in fact find a pattern in x and y).

(1 - sqrt 2)(1 + sqrt 2) = -1.

iluvmaths, you are very impressive!
EDIT: Hypothesis; [Sqrt(r) - 1]^n = Subtraction of 2 sq roots, difference between the integers in the sq roots is (r-1)^n.

now this is beautiful... i realised that sqrt2 -1 is equal to x +ysqrt2 but (sqrt2 +1)^n = x-ysqrt2 is wonderful.. but.. i didnt realise that (sqrt2 -1)^n(sqrt2 +1)^n = (-1)^n... wonderfull Dfranklin.. this is why u went to cambridge...

now this impressive

It was a typo, he meant (1 - Sqrt2)^n = x + ySqrt2, and similarly (1 + Sqrt2)^n = x - ySqrt2.
As (1 - Sqrt2)(1 + Sqrt2) = -1, then (1 - Sqrt2)^n.(1 + Sqrt2)^n = (-1)^n and so on ...

well yeah, btw, we need to be proving the hypothesis... its just beautiful

iluvmaths, you are very impressive!
EDIT: Hypothesis; [Sqrt(r) - 1]^n = Subtraction of 2 sq roots, difference between the integers in the sq roots is (r-1)^n.

Note that:
$\sqrt 2 - 1 = \sqrt 2 - \sqrt 1$
$(\sqrt 2 - 1)^2 = \sqrt 9 - \sqrt 8$
$(\sqrt 2 - 1 )^3 = \sqrt{50} - \sqrt{49}$
$(\sqrt 2 - 1 )^4 = \sqrt{289} - \sqrt{288}$
And so on.
So we can hypothesise that, for any positive integer $n$, $(\sqrt 2 - 1 )^n$ can be expressed as the difference of the square roots of two consecutive integers.
How would we prove this?
I've tried things but I can't seem to get anywhere. I don't think I can use induction for this because the consecutive rooted integers seem to have no pattern between them as n increases (at least I haven't spotted it yet). Then again, it is fairly late so I'm not at my sharpest.
Thanks in advance.