Prove the existence of the number $n$ that satisfies a property.

Question

Problem

Prove that there is a $n \in$ $ \mathbb{N} $ such that the equation has at least 2022 solutions.

$x+y+\sqrt{xy}=n$ ,with $x,y\in$ $ \mathbb{N}$

Attempt

My first attempt was to fix $x$ to $0, 1, 2$ and try to simplify the problem a bit but it would not work. I continued and wrote a c++ code in order to vizualize the expressions for a $n$ up to a large number, but it did not help me visualize any pattern.

I thought about considering the function $f(x,y) = x+y+\sqrt{xy}$. We want to find a positive integer $n$ such that $f(x,y) = n$ has at least 2022 solutions in positive integers $x$ and $y$.

Let $k$ be a positive integer. We claim that if we choose $n$ such that $n\geq k+2$ and $n^2-k\geq 4n$, then the equation $f(x,y) = n$ has at least $k+1$ solutions in positive integers $x$ and $y$.

And, here I am completely stuck! Any help or idea would be apprecieted!

Answer 1

If you focus on just finding an $n$ with many solutions (obviously $2022$ in the problem statement is just a placeholder for 'how many you want'), this problem is easy if you go the other way around: Find parameterized solutions $(x,y,n)$, then choose the parameters such that the same $n$ appears as often as you want.

Since $\sqrt{xy}$ needs to be an integer, given any $x$, choosing $y=x$ seems to be the most immediate choice to make that true. That means $n=x+x+\sqrt{x\times x}=3x$. This means $(x,x,3x)$ is a solution triple.

That means any $n$ divisible by $3$ has at least $(x,y)$ solution, namely $x=y=\frac{n}3$. But for a given $n$, this is only one solution.

But the method can be easily generalized by seeing that $y=k^2x$ for $k \in \mathbb N, k \ge 1$ also leads to a natural $\sqrt{xy}$ (the above is just $k=1$).

In that case, we get $n=x+k^2x+\sqrt{x\times k^2x} = x+k^2x+kx=(1+k+k^2)x.$

That means for each positive $x$ and $k \in \mathbb N$,

$$(x, k^2x,(1+k+k^2)x) \tag{1}$$ is a solution triple.

That means any $n$ divisible by $1+k+k^2$ has a solution of the form $(1)$. That means you 'only' need to find an $n$ that is divisible by $1+k+k^2$ for $2022$ different values of $k$, say $k=1,2,\ldots,2022$. It will be very large, but it certainly exists, let's call it $n_{2022}$.

For different $k$ the $x$ and $y$ values generated by $(1)$ are all different, because for example $x=\frac{n_{2022}}{1+k+k^2}$ and the denominator is a strictly increasing function of $k$ for positive $k$, so all the $x$ values are different.

Answer 2

Hint: Prove by induction on the (at least) number of solutions.
If you have an $n$ with at least $k$ solutions of the form $(x_i, y_i)$, then $an$ also has at least $k$ solutions of the form $(ax_i, ay_i)$.
So, for a specially selected $a$, if we can find another solution in which $x_i$ is not a multiple of $a$, then this gives us at least $k+1$ solutions.

EG Take $ a = 1 + n + n^2$, then $ x = n, y = n^3$ satisfies $ x + y + \sqrt{xy} = an$.
So $ (n, n^3, an)$ is our $k+1$st solution.

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In fact, modifying that idea, if we have solutions $(x_i, y_i , n_i)$ with $\frac{x_i}{y_i}$ distinct, then we can combine them to get the distinct pairs $X_i = x_i \prod_{j\neq i} n_j, Y_i = y_i \prod_{j\neq i} n_j$ which satisfy $X_i + Y_i + \sqrt{X_iY_i} = \prod n_i$.
EG You can pick any squares $x_i, y_i$, as long as they satisfy the distinct condition.

As an explicit construction, with $(x_i, y_i, n_i) = (1, i^2, 1+i+i^2)$, we get that $n = \prod_{j=1}^{2022} (1 + j + j^2)$ has the following 2022 solutions:

$$x_i = \frac{1}{ 1 + i + i^2} \prod (1+j+j^2), \quad y_i = \frac{i^2}{1+i+i^2} \prod ( 1 + j + j^2) .$$

Answer 3

This problem is actually quite solvable exactly in terms of a rational parameterization. It just takes a bit of work to get there. Writing

$n - (x+y) = \sqrt{xy} $

$n^2 -2n(x+y) + x^2 +2xy + y^2 = xy $

$n^2 -2n(x+y) + x^2 +xy + y^2 = 0 $

So we have a quadratic function in terms of x and y. We rewrite this expression in the following convenient manner

$x^2 + 2( \frac{y}{2} -n )x + y^2 -2ny + n^2 =0 $

Completing the square in terms of $x$, that is adding and subtracting $ (\frac{y}{2} - n)^2$ to the left hand side, we can simply the equation in terms of

$(x+ \frac{y}{2} - n)^2 + \frac{3}{4}y^2 - ny =0 $

Setting $u = x+ \frac{y}{2} - n$, can find a nice expression

$u^2 + \frac{3}{4}y^2 - ny =0 $

So now we can find a convenient way to express this for the purpose of analyzing the equations

$\frac{ny}{ u^2 + \frac{3}{4} y^2 }{ny} = 1 $

The function

$ F(u, y ) = \frac{ u^2 + \frac{3}{4} y^2 }{ny}$

is homogenous of order 1, so it can be written in the form $ F = uf(\frac{y}{u}) $ for example. So it is reasonable to conclude that setting $ y = t u$, we can solve for $u$ in terms of the parameter $t$ in the given equation. Therefore,

$u^2 + \frac{3}{4} t^2 y^2 - nt u =0 $

$u[ (1 + \frac{3}{4} t^2 ) u - nt ] =0$

So either $u=0$, or

$u = \frac{4nt}{4 + 3 t^2} $; $ y = \frac{4nt^2}{4 + 3 t^2}$

You can now solve for $x$ in terms of $u$ and $y$ and to find a rational parameterization for $x$ and $y$ which you can simplify to a nice form and check that this is satisfies the initial equation. After which, it's simply a question of which rational numbers $t$ yield natural $x$ and $y$ since $x$ and $y$ are guaranteed to be rational in this parameterization for any rational number $t$.