Another Good Number?

Lemma 1:- If $l^{p_n}k$ is an integer for a monotonically increasing sequence $<p_n>$ with $p_1 = 1$ and a non-zero integer $k$ , then $l$ is an integer.

Proof :- $l$ can't be irrational, otherwise it would imply $lk$ is irrational, contradiction. Then, for the sake of contradiction, suppose $l$ is a non integral rational number $\frac{p}{q}$ where $p,q$ are integers, $q \not = 1$ and $\gcd(p,q) =1$ . Then $l^{p_n}k = \frac{p^{p_n}k}{q^{p_n}}$ . Since $\gcd(p^n,q^n) =\gcd(p,q) =1$ , we have $q^{p_n} | k \implies q^{p_n} \leq k$ . But this is not possible as $q^{p_n}$ is increasing. Therefore, our suppostion is wrong and $l$ is an integer.

Now, $(a^n - b^n)^2 + a^{2n} - b^{2n} = 2a^n(a^n-b^n)$ implies $2a^n(a^n-b^n)$ is an integer.

Also, $[(2a^n)^k(a^n-b^n)]^2 + 2^k[(2a^{2n})^k(a^{2n}-b^{2n})] = (2a^n)^{2k +1}(a^n - b^n)$ implies that if $(2a^n)^k(a^n - b^n)$ is an integer, then $(2a^n)^{2k +1}(a^n - b^n)$ is an integer for natural $n$ . Since the case $k =1$ is true, we can conclude that $(2a^n)^{t_n}(a^n - b^n)$ is an integer for a monotonically increasing sequence $<t_n>$ with $t_1 =1$

Then, by Lemma 1, $2a^n$ must be an integer for all natural numbers $n$ . Applying Lemma 1 again,we see $a$ must be an integer.

Now since $a -b$ and $a$ both are integers, $b$ must also be an integer. Hence, if $a^n - b^n$ is an integer for all natural $n$ , then $a$ and $b$ must be integers.

a-b is an integer, then a=x+m, b=y+m, with x, y are integer and 0<=m<=1

a^{2} - b^{2} = (a-b) \times (a+b) is an integer => a+b is an integer also => m=0.5 or 0

a^{3} - b^{3} = (a-b) \times ( a^{2} + b^{2} + a \times b ) is integer also => m must be 0 => a and b must be integer

Just a comment. Are we saying that sqrt (2) is not a distinct real number? Because I was thinking (sqrt (3))^2-(sqrt(2))^2=1. I think we should use the term "rational" instead of "distinct".