show this diophantine equation has at least is $3n+3\lfloor \frac{n+1}{3}\rfloor+1$ postive integer solution

Question

For any postive integer $n\ge 4$, let $s(n)$ denote the number of ordered pairs $(x,y,z)$ of positive integers for which $$\color{red}{xy+yz+xz=n(x+y+z)}$$ show that $$s(n)\ge 3n+3\lfloor \dfrac{n+1}{3}\rfloor+1$$

Thanks

Answer 1

In this case, for the equation.

$$xy+xz+yz=n(x+y+z)$$

Choose and define yourself the number $z$. Then substitute in the formula and place into factors.

$$ab=n^2-nz+z^2$$

And then the remaining solution will look like this.

$$x=b+n-z$$

$$y=a+n-z$$

Answer 2

Here's an approach to construct the necessary quantity of solutions which uses the ideas suggested by individ:

So we fix some $x=k$ where $1\le k \le 10$ and obtain the equation $$(y-n+k)(z-n+k)=n^2-kn+k^2$$

Now, we are searching for a factorization of the RHS. There is one obvious factorization which works for all $k$: $$n^2-kn+k^2=1 \cdot (n^2-kn+k^2)$$

This gives the solution $(x,y,z)=(k,n-k+1,k^2-nk+n^2+n-k)$.

Let's look at these solutions for $k$ running from $1$ to $\lceil \frac{n}{2} \rceil$. Here, it's easy to verify that for $n \ge 4$ all the 3 numbers are distinct and we have $x<y<z$.

Since we can have 6 permutations of each such solution, this gives $6 \cdot \lceil \frac{n}{2} \rceil \ge 6 \frac{n}{2}=3n$ solutions.

Also, we can find the solution $(x,y,z)=(n,n,n)$. So we are just left to find $3\lfloor \frac{n+1}{3} \rfloor$ more solutions.

Therefore, we note that if we choose $k$ such that $n+k$ is divisible by 3 i.e. $k \equiv -n \mod 3$ then $n^2-kn+k^2 \equiv 3n^2 \equiv 0 \mod 3$ i.e. $n^2-kn+k^2$ is divisible by 3.

Hence, we can here construct solutions with $x=k, y=n+3-k$ and solve the remaining equation to $z=\frac{n(x+y)-xy}{3}=\frac{(n-k)(n+3)+k^2}{3}$. Since we must choose $k \le n+2$ there are at least $\lfloor \frac{n+2}{3} \rfloor$ possible values of $k$. This would give $6 \lfloor \frac{n+2}{3} \rfloor$ solutions for $(x,y,z)$ (again using the permutations) but it's easy to see that each solutions is counted twice so we obtain at least $3 \lfloor \frac{n+2}{3} \rfloor$ additional solutions.

So after checking that the three sets of constructed solutions are indeed disjoint (which is a bit tedious but essentially just casework and rough estimations) we see that we have indeed constructed at least $$3n+3\lfloor \frac{n+2}{3} \rfloor +1$$ distinct solutions $(x,y,z)$ which is clearly sufficient.