×

# How to prove???

Let $$x$$ and $$y$$ be real numbers such that $$x+y, x^{2}+y^{2}, x^{3}+y^{3}$$ and $$x^{4}+y^{4}$$ are integers. If $$n$$ is a natural number, prove that $$x^{n}+y^{n}$$ is an integer.

Help would be greatly appreciated!

Note by Victor Loh
2 years, 7 months ago

Sort by:

The key step is to show that $$xy$$ is an integer, and then use (strong) Induction on the algebraic identity

$x^{n+1} + y^{n+1} = (x+y) (x^n + y^n) - xy ( x^{n-1} + y^{n-1} ).$

Note that you have to use the $$x^4 + y^4$$ condition, as just the first three are not strong enough. For example, using $$x = \frac{\sqrt{2}}{2}, y = - \frac{ \sqrt{2} } {2}$$, gives us $$x+y = 0, x^2 + y^2 = 1, x^3 + y^3 = 0$$ but $$x^4 + y^4 = \frac{1}{2}$$ (and the higher even powers clearly do not yield integers). Staff · 2 years, 7 months ago

This was my method! · 2 years, 7 months ago

Hi Calvin

When I was doing the problem I also got stuck with xy. But can u further elaborate on the later part of your hint? Thanks · 2 years, 7 months ago

I haven't found how to get $$xy$$ is an integer; I only got $$xy$$ is half an integer. I'm still figuring out how to get past that. · 2 years, 7 months ago

To get that $$xy$$ is half an integer, you only used the first 2 conditions. Namely,

$2xy = (x+y)^2 - (x^2 + y^2 ) \in \mathbb{N}.$

Use the "hint" that you must use the 4th condition. Somehow or other, it must come into play (and it does, in a very similar manner). Staff · 2 years, 7 months ago

If $$x$$ and $$y$$ are irrational then there is no way that the initial integer conditions hold, so let us write $$x = \dfrac{a}{m}$$ and $$y = \dfrac{b}{n}$$ where $$gcd(a,m) = gcd(b,n) = 1$$. Let us assume that $$m \ne n$$, ie, there is some prime factor in $$m$$ which is not in $$n$$. Then $$x + y = \dfrac{a}{m} + \dfrac{b}{n} = \dfrac{an+bm}{mn}$$. Considering the numerator, $$an + bm \equiv an \pmod{m}$$ but there is a prime factor in $$m$$ which is not in $$n$$ and $$gcd(a,m) = 1$$ so $$an \not\equiv 0 \pmod{m}$$ ie. the numerator can't be divisible by $$m$$ and so $$x + y$$ is not an integer. Therefore our assumption was wrong so $$x$$ and $$y$$ have the same denominator. So we will rewrite them as $$x = \dfrac{a}{m}$$ and $$y = \dfrac{b}{m}$$ where again $$gcd(a,m) = gcd(b,m) = 1$$.

We have that $$x + y$$ is an integer so $$a + b \equiv 0 \pmod{m}$$ so $$a \equiv -b \pmod{m}$$. Squaring both sides of the congruence gives $$a^2 \equiv b^2 \pmod{m}$$. But we also have that $$x^2 + y^2$$ is an integer so $$a^2 + b^2 \equiv 0 \pmod{m}$$ so $$a^2 \equiv -b^2 \pmod{m}$$ (in fact $$a^2 + b^2 \equiv 0 \pmod{m^2}$$ but that is less useful to us). Combining these congruences gives $$b^2 \equiv -b^2 \pmod{m}$$ so $$2b^2 \equiv 0 \pmod{m}$$. But $$gcd(b,m) = 1$$, so either $$m = 1$$ or $$m = 2$$.

If $$m = 1$$ then the result is trivial (since $$x$$ and $$y$$ themselves are integers ). If $$m = 2$$ then $$a$$ and $$b$$ are odd, so $$a^2 + b^2 \equiv 2 \pmod{4}$$ but $$x^2 + y^2 = \dfrac{a^2 + b^2}{4}$$ so that means that $$x^2 + y^2$$ is not an integer.

So your initial set of conditions only hold if $$x$$ and $$y$$ are integers themselves, following which the result is obvious. · 2 years, 7 months ago

Oh, in fact, I can produce a counterexample to your claim: $$x = \sqrt{2}, y = -\sqrt{2}$$ satisfies the conditions of the problem but not your claim that $$x,y$$ are integers. · 2 years, 7 months ago

Another statement which is not true is "Let us assume that $$m\neq n$$, i.e., there is some prime factor in $$m$$ which is not in $$n$$." The IE part doesn't necessarily follow from the assumption. For example, take $$m = 2, n = 4$$. Note that with these values, we have $$an \equiv 0 \pmod{m}$$, which contradicts your conclusion. This could be fixed though, with better bookkeeping of the variables. Staff · 2 years, 7 months ago

I was considering WLOG ( m > n ) and so either a prime factor or prime exponent are different which gives the same result · 2 years, 7 months ago

If $$x$$ and $$y$$ are irrational then there is no way that the initial integer conditions hold

How exactly? · 2 years, 7 months ago

Very nice solution! · 2 years, 7 months ago

We use the identity $$x^n+y^n=(x+y)(x^{n-1}+y^{n-1})-xy(x^{n-2}+y^{n-2})$$.

First we prove that $$xy$$ is an integer. Note that $$2xy=(x+y)^2-(x^2+y^2)\in \mathbb{Z}$$ and $$2x^2y^2=(x^2+y^2)^2-(x^4+y^4)\in \mathbb{Z}$$. From the first equation we see that $$2xy=m$$ for some integer $$m$$, so $$xy=\frac{m}{2}$$. From the second equation we have $$2x^2y^2=n$$ for some integer $$n$$. Plugging $$xy=\frac{m}{2}$$ yields $$2\left(\frac{m^2}{4}\right)=n\Leftrightarrow \frac{m^2}{2}=n\in \mathbb{Z}$$. So $$2\mid m$$. Hence $$xy=\frac{m}{2}\in \mathbb{Z}$$, as desired. //

Now, let $$p(n)=\text{true}\Rightarrow x^n+y^n\in \mathbb{Z}$$. By the identity, $$p(1), p(n-1), xy, p(n-2)\Rightarrow p(n)$$. So $$p(1), p(4), xy, p(3)\Rightarrow p(5)$$; $$p(1), p(5), xy, p(4)\Rightarrow p(6)$$, and by a straight-forward strong induction on $$n$$, $$p(n)=\text{true}\ \forall n\in \mathbb{N}$$. $$\Box$$ · 2 years, 6 months ago

I would greatly appreciate help too.

I am serious. · 2 years, 7 months ago

You are totally 44 · 2 years, 7 months ago

you are totally 18 · 2 years, 7 months ago

You are totally 14 · 2 years, 7 months ago

Come on my age is wrong... I am victor's classmate. Our homework for math was this question. So yeah, I am 14. · 2 years, 7 months ago

13. · 2 years, 7 months ago