# Let $k \geq 2$ be an integer. Let $x$ and $y$ be positive integers. Show that $x^k- y^k > 2$.

Let $k \geq 2$ be an integer. Let $x$ and $y$ be positive integers. Show that $x^k- y^k > 2$.

I'm a little confused by this because this is the only information given. The book I'm using doesn't specify whether or not $x$ needs to be greater than $y$, so couldn't you have $2^2 - 3^2 = -5$?

Also, I just don't know how to approach this problem.

• Presumably $x>y$. Try factoring $x^k-y^k$. – carmichael561 Apr 3 '16 at 23:25
• Or... let $x = y + n; n>0$ then prove $(x+n)^k - x^k >2$. – fleablood Apr 3 '16 at 23:36
• Although I have to get irked that it doesn't stipulate x > y. If $x \le ys this is obviously not true. – fleablood Apr 3 '16 at 23:38 ## 2 Answers If there is not the condition that$x>y$then your counterexample is a good one and immediately yields that the statement is false in general. If we add the additional constraint that$x>y$, then we know that$x^k-y^k$is a positive integer. (if you do not wish to take this as fact and wish to prove this statement, note that integers are closed under multiplication and$x>y\Rightarrow x^k>y^k$) Next, we note that we can factor$x^k-y^k$.$x^k-y^k = (x-y)(x^{k-1}+x^{k-2}y+x^{k-3}y^2+\dots+xy^{k-2}+y^{k-1})$Since$k\geq 2$, and$x>y>0$are both integers, we know that$x\geq 2$as well. The above factorization then has that both parts are positive integers and furthermore that the righthand part is at least$3$since there are at least two positive integer terms in the summation, with at least one of them a multiple of two. Thus$x^k-y^k>2$for all positive integers$x>y$and$k\geq 2$. Since$x,y\in\Bbb N$we may as well assume$x>y\iff x\ge y+1$. But then $$x^k-y^k\ge x^k -(x-1)^k.$$ Set$z+1= x$then it is enough to show$(z+1)^k-z^k>2$. $$(z+1)^k-z^k = \sum_{n=0}^{n-1}{k\choose n} z^n> {k\choose 1}$$ here the last inequality follows by positivity of$z$. But${k\choose 1} = k$and$k\ge 2\$.

So we have

$$x^k-y^k\ge x^k-(x-1)^k > {k\choose 1} \ge 2.$$