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!

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!

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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). – Calvin Lin Staff · 3 years, 2 months ago

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– Wei Jie Tan · 3 years, 2 months ago

This was my method!Log in to reply

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 – Victor Loh · 3 years, 2 months ago

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– Ivan Koswara · 3 years, 2 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.Log in to reply

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

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). – Calvin Lin Staff · 3 years, 2 months ago

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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. – Josh Rowley · 3 years, 2 months ago

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– Ivan Koswara · 3 years, 2 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.Log in to reply

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. – Calvin Lin Staff · 3 years, 2 months agoLog in to reply

– Josh Rowley · 3 years, 2 months ago

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

How exactly? – Ivan Koswara · 3 years, 2 months ago

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– Cody Johnson · 3 years, 2 months ago

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 \(n\), \(p(n)=\text{true}\ \forall n\in \mathbb{N}\). \(\Box\) – Arkan Megraoui · 3 years, 1 month ago

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I would greatly appreciate help too.

I am serious. – Yuxuan Seah · 3 years, 2 months ago

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– Wei Jie Tan · 3 years, 2 months ago

You are totally 44Log in to reply

– Victor Loh · 3 years, 2 months ago

you are totally 18Log in to reply

– Wei Jie Tan · 3 years, 2 months ago

You are totally 14Log in to reply

– Yuxuan Seah · 3 years, 2 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.Log in to reply

– Wei Jie Tan · 3 years, 2 months ago

13.Log in to reply

STATE YOUR SOURCE – Wei Jie Tan · 3 years, 2 months ago

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