Are there integer solutions such that the sum of the roots of x and y equal the root of a prime p?

Question

How could one prove whether or not integer pairs (x, y) exist such that $\sqrt{x} + \sqrt{y} = \sqrt{p}$, where p is prime? `

Answer 1

Using elementary methods. We'd have

$$x + 2\sqrt{xy} + y = p$$

and hence

\begin{align} x + y - p = -2\sqrt{xy} &\implies x^2+y^2+p^2+2xy-2xp-2yp = 4xy \\&\iff (x^2-2xy+y^2) + p^2= 2p(x+y) \\&\iff (x-y)^2 +p^2 = 2p(x+y)\tag{1} \\&\iff (x-y)^2 +2p(x-y) +p^2 = 2p(x+y) + 2p(x-y)\tag{2} \\&\iff (x-y+p)^2 = 4px. \end{align}

This implies $x-y+p = \pm 2\sqrt{px}$, so $x = pa^2$ for some integer $a$.

At $(2)$, we added $2p(x-y)$ to both sides of equation $(1)$. If we instead had subtracted $2p(x-y)$, we'd get

\begin{align} x + y - p = -2\sqrt{xy} &\implies (x-y)^2 -2p(x-y) +p^2 = 2p(x+y) - 2p(x-y)\tag{3} \\&\iff (x-y-p)^2 = 4py. \end{align}

Like before, we obtain that $x-y-p = \pm 2\sqrt{py}$, and hence $y = pb^2$ for some integer $b$.

The initial equation can then be rewritten:

$$|a|\sqrt p + |b|\sqrt p = \sqrt p \iff |a| + |b| = 1.$$

This can happen only when $|a| = 1$ and $b=0$ or vice versa, that is, either $x = p$ and $y=0$ or $x=0$ and $y=p$.

Answer 2

For any $p$ you could take, $(0,p)$, or $(p,0)$, but let's assume $x$ and $y$ are nonzero. Then $$x+2\sqrt {xy}+y=p$$ which means $2\sqrt{xy}=p-x-y$ and thus $$4xy=(p-x-y)^2=p^2-2p(x+y)+(x+y)^2$$ so modulo $p$, $4xy\equiv (x+y)^2$, and thus $0\equiv x^2-2xy+y^2=(x-y)^2$. But then $p$ divides $x-y$ which is impossible since $x-y$ is an integer strictly between $-p$ and $p$.

Answer 3

Assume $x\gt y$. Let $\sqrt x - \sqrt y=a$ and multiply this by $\sqrt x +\sqrt y=\sqrt p$ to obtain $$x-y=a\sqrt p$$ $x-y$ is an integer $n$ and $a=\frac np\sqrt p=r\sqrt p$ is a rational multiple of $\sqrt p$

So $2\sqrt x=(1+r)\sqrt p$ and squaring this we find that $r$ is an integer and $p|x$. Similarly $p|y$.

If $x=pt, y=ps$ we are reduced to $\sqrt t+\sqrt s=1$ and since $t\gt s$ we have $t=1, s=0$ as the only solution in integers.

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For $r$ is an integer $4x=\frac {(p+n)^2}{p^2}p$ and $4px=(p+n)^2$ so $p|(p+n)$ whence $p|n$

Answer 4

A slightly less elementary approach. Consider the polynomial whose roots are the pairwise sums of the roots of the minimal polynomials of $\sqrt x$ and $\sqrt y$: $$(z - \sqrt x - \sqrt y)(z - \sqrt x + \sqrt y)(z + \sqrt x - \sqrt y)(z + \sqrt x + \sqrt y) \\ = ((z - \sqrt x)^2 - y) ((z + \sqrt x)^2 - y) \\ = (z^2 + x - y - 2z\sqrt x) (z^2 + x - y + 2z\sqrt x) \\ = (z^2 + x - y)^2 - 4z^2x \\ = z^4 - 2(x+y) z^2 + (x-y)^2$$ If $\sqrt p$ is a root then its minimal polynomial $z^2 - p$ is a factor. Since we have two monic polynomials, that implies that $p$ is a factor of $(x-y)^2$. The tail of Elliot's argument wraps it up.

We could instead have gone on to do a long division, finding that the remainder is $$p^2 - 2(x+y)p + (x-y) ^2 =0$$ as in Fimpellizieri's answer.