How to prove???

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.

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).

I would greatly appreciate help too.

I am serious.

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

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