Show that for any integer n there exists an integer m such that:?

By taking inverses you would have

(sqrt(2) + 1)^n = sqrt(m) + sqrt(m-1), so that

2*sqrt(m) = (sqrt(2)-1)^n +(sqrt(2)+1)^n

and the expression of m follows. I guess this is what you called a/ right?

Can you use induction? If you can, then, simply by multiplying by

(sqrt(2)-1)^2 + (sqrt(2)+1)^2,

you see that the sequence m_n has the form

m_(n+1) = 6*m_n - m_(n-1) - 2

and so you get 2,9,50,289....... etc

Incidentally, except for the (-2) this is the induction formula which yields the hypotenuses (w) in the Pythagorician triplets of the form (u,u+1,w)

If they have an elementary way to see that (sqrt(2)-1)^n and (sqrt(2)+1)^n are of the form

b sqrt(2) - a and b sqrt(2)+a when n is odd and

- b sqrt(2) + a and b sqrt(2)+a when n is even

then by multiplying one gets that

2b^2 - a^2 is always + 1 or -1.

So m = 2b^2 or a^2, depending on the parity.

a million) if A is wonderful then A² is wonderful say, M = ok+a million and N = ok (ok : any beneficial integer) M²-N² = (ok+a million)² - ok² = 2k+a million (= any beneficial wonderful integer) so there are constantly 2 integers M and N at the same time with M²-N² : is any wonderful integer examples A = 15 = 2*7+a million ==> ok = 7 15 = 8² - 7² A² = 15² = 225 = 2*112 + a million ==> ok = 112 15² = 113² - 112² 2) if A is even then A² is even say, M = ok+a million and N = ok-a million (ok : any beneficial integer) M²-N² = (ok+a million)² - (ok-a million)² = 4k =2²ok A = 2B (B : integer) A² = 2²B² = 2²ok ==> ok = B² to fulfill the situation examples A = 18 = 2*9 ==> A² =18² = 2²*9² ==> ok = 80 one 18² = 80 two² - 80² A = 256 = 2*128 ==> A² =256² = 2²*128² ==> ok = 128²=16364 256² = 16365² - 16363² then for any integer A, we are able to discover a minimum of two integers M,N at the same time with A² = M² - N² ======================================...

Okay, this is an re-edited answer. All powers of (√2 - 1) are of the form A + B, where A and B are either one of the two, one of which at least contains the radical √2:

(1/2)( (1+√2)^x - (1-√2)^x )

(1/2)( (1+√2)^x + (1-√2)^x )

Proving that A² = B² ± 1 then proves this conjecture. Unfortunately, I am running out of time to explain fully.

Now I am not too sure about how to answer this, but here is my attempt:

(√2 - 1)^n = √m - √(m-1)

log (√2 - 1)^(√2 - 1)^n = log (√2 - 1)^(√m - √(m-1)

n = log (√2 - 1)^(√m - √(m-1)

It is then rewritten with natural log:

n = ln(√m - √(m-1)) / ln(√2 - 1)

Thus, I think it is proven that for any value "n", you can generate a real number consisting of the integer "m" (sense the bottom produces a constant). So was this what you were looking for?