Showing that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n$

Question

Show that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n.$

I'm not very certain how to proceed on this problem. I know $x^2+y^2=(x+y)^2-2xy,$ so we essentially have $x+y=3n, (x+y)^2-3xy=7n$ for positive integers $x, y, n.$ However, this doesn't really help. I've also tried writing it as a fraction and doing some algebraic manipulations, but I haven't gotten anywhere either. May I have some help? Thanks in advance.

Answer 1

We are given

$$x + y = 3n, \; x^2 + y^2 - xy = 7n \tag{1}\label{eq1A}$$

As you showed,

$$\begin{equation}\begin{aligned} (x + y)^2 - 3xy & = 7n \\ 9n^2 - 3xy & = 7n \\ 3xy & = 9n^2 - 7n \\ xy & = 3n^2 - \frac{7n}{3} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

From this, plus the first part of \eqref{eq1A}, then using Vieta's formula's gives that $x$ and $y$ are the roots of the quadratic equation

$$z^2 - (3n)z + \left(3n^2 - \frac{7n}{3}\right) = 0 \tag{3}\label{eq3A}$$

The quadratic formula gives

$$\begin{equation}\begin{aligned} z & = \frac{3n \pm \sqrt{9n^2 - 4\left(3n^2 - \frac{7n}{3}\right)}}{2} \\ & = \frac{3n \pm \sqrt{-3n^2 + \frac{28n}{3}}}{2} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Using that $n$ is positive, then factoring the part in the square root above (i.e., the discriminant), and requiring it to be non-negative, gives

$$-3n^2 + \frac{28n}{3} = n\left(-3n + \frac{28}{3}\right) \implies -3n + \frac{28}{3} \ge 0 \implies n \le \frac{28}{9} = 3 + \frac{1}{9} \tag{5}\label{eq5A}$$

We have \eqref{eq1A} indicating $n$ must be either an integer or a rational value of the form $\frac{m}{3}$ for some positive integer $m$ where $3 \not\mid m$. For the latter case, though, the second part of \eqref{eq1A} would not be an integer. Thus, $n$ is an integer, with \eqref{eq2A} showing it must be a multiple of $3$.

Next, the upper limit in \eqref{eq5A} shows $n = 3$ is the only possible solution, with \eqref{eq4A} giving $z = \frac{9 \pm 1}{2}$, i.e., $z = 4$ or $z = 5$. Thus, we get $(x,y) = (4,5)$ or $(5,4)$.

Answer 2

Square the first equation and then subtract the second from it to obtain $$9n^2-7n=3xy\tag1$$ In particular, $3$ divides $n,$ so we can write $n=3k$ for $k\ge1,$ and transform (1) into $$27k^2-7k=xy\tag2$$ and then into $$27k^2-7k=x(9k-x)\tag3$$ If $k=1,$ then the desired result follows from (3). Let $L$ and $R$ denote the left and right sides of (3). If $k\ge2,$ then $$ L= 21k^2+6k^2-7k\gt21k^2\gt20.25k^2\ge R$$

Answer 3

Because $x$ and $y$ are exchangeable, I represent them as $$ x=m+d \qquad y=m-d $$ Then $$ x+y=3n \qquad \to \qquad 2m = 3n \qquad \to \qquad 7n= 14/3m \tag 1$$ $$ x^2+y^2-xy = 7n \qquad \to \\ 2m^2+2d^2 - (m^2-d^2)= 7n \qquad\to \\ m^2+3d^2 = 14/3m \qquad\to \tag 2 $$ $$ m^2-14/3m + (7/3)^2 = (7/3)^2-3d^2 \qquad\to $$ $$ m= 7/3 \pm \sqrt{ (7/3)^2-3d^2} \tag 3 $$ The integer resp half-integer values of the term $7/3 \pm \sqrt{ (7/3)^2-3d^2}$ can be enumerated for $d \in \{0,1/2,1,3/2,...\}$ and all for $d \gt 1 $ become imaginary. From that only $d=0$ and $d=1/2$ are integer or half integer and lead to the solutions:

d    m                x=m+d  y=m-d
--------------------------------------- for 7/3 + sqrt(...)
0    4.66666666667    4+2/3  4+2/3
0.5  4.50000000000    5      4           <---- the single integral solution
1    3.89680525327    <fractional> 
--------------------------------------- for 7/3 - sqrt(...)
0    0                  "trivial"
0.5  0.166666666667   <fractional>
1    0.769861413392   <fractional>

Conclusion: The only integer-solutions are $(x,y)=(y,x)=(5,4)$