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!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe 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).

Log in to reply

This was my method!

Log in to reply

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

Log in to reply

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.

Log in to reply

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).

Log in to reply

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.

Log in to reply

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.

Log in to reply

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.Log in to reply

I was considering WLOG ( m > n ) and so either a prime factor or prime exponent are different which gives the same result

Log in to reply

How exactly?

Log in to reply

Very nice solution!

Log in to reply

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 \(n\), \(p(n)=\text{true}\ \forall n\in \mathbb{N}\). \(\Box\)

Log in to reply

I would greatly appreciate help too.

I am serious.

Log in to reply

You are totally 44

Log in to reply

you are totally 18

Log in to reply

Log in to reply

Log in to reply

Log in to reply

STATE YOUR SOURCE

Log in to reply