Another Non-Homogenous Diophantine Equation

This is kind of an ugly solution. Maybe there is something better?

Anyway, it's clear that $y$ must be even. Writing $y=2z$ and simplifying gives $2x^3+2z^3 = 3xz+1.$ Set $z=a-x$ and rewrite to get $(6a+3)x^2-(6a^2+3a)x+(2a^3-1) = 0.$ The discriminant of this quadratic polynomial is $(2-a)(2a+1)(6a^2+3a+6).$ This is nonnegative for $-1/2 \le a \le 2,$ since the quadratic factor is always positive. But $a$ is an integer, so it remains to check $0,1,2.$

When $a=0$ we get $3x^2-1=0,$ no integer solutions.

When $a=1$ we get $9x^2-9x+1 = 0,$ no integer solutions.

When $a=2$ we get $15x^2-30x+15 = 0,$ with the unique solution $x=1.$ This gives $z=1$ and $y=2.$ So the unique solution is $(x,y) = (1,2)$ and the answer is $1+1+2=\fbox{4}.$

$8x^3 - 4 = y(6x-y^2) \\ 8x^3 - 4 = 6xy - y^3 \\ 8x^3 + y^3 = 2\left( 4x^3 + \cfrac{y^3}{2} \right) = 2(3xy + 2) = 6xy + 4$

Since $y^3 \ \text{mod} \ 2 = 0 , \implies y = 2k$

$4x^3 + 4k^3 = 6kx + 2 \\ x^3 + k^3 = \cfrac{3kx + 1}{2}$

Since $3kx + 1 \ \text{mod} \ 2 = 0 , \implies kx \ \text{is odd and hence,} \ k = 2p+1 , x = 2q + 1$

$8(p^3 + q^3) + 12(p^2 + q^2) + 6(p+q) + 2 = 6pq + 3(p+q) + 2 \\ 2pq + 2pq + 2pq = 8(p^3 + q^3) + 12(p^2 + q^2) + 3(p+q) \\ p = q = 0 \ \text{is the only possible integral solutions for p and q} \\ k = 1 , x = 1 \\ \implies x = 1 , y = 1 \\ \therefore n + \sum_{i = 1}^{n} \left( x_i + y_i\right) = 1 + 1 + 2 = \boxed{4}$

We start by solving the equation $2(a^3 + b^3) = 3ab + 1$ for integers $a,b$ . It is clear that neither $a$ nor $b$ can be zero, and so there are two cases to consider:

(1) If $ab > 0$ then $3ab+1 > 0$ , and hence $a,b > 0$ . Without loss of generality we can assume that $a \ge b \ge 1$ . Then $2a^3 \; \le \; 2(a^3 + b^3) \; = \; 3ab + 1 \; \le \; 3a^2 + 1$ so that $2a^3 - 3a^2 + 1 \le 0$ , which implies that $a < 2$ . Thus the only possible solution in this case is $a=b=1$ .

(2) If $ab < 0$ then we can write $b = -c$ and consider the equation $2(c^3 - a^3) = 3ac - 1$ where $c \ge a \ge 1$ . But then $2(c-1)^3 \; \le \; 2(c-1)^3 + 3(c-1)^2 \; = \; 2c^3 - 3c^2 + 1 \; = \; 2a^3 + 3ac - 3c^2 \; \le \; 2a^3$ and hence $a \ge c-1$ . Thus either $a=c-1$ or $a=c$ . The option $a=c-1$ is only possible if $c^2 - c + 1 = 0$ , while the option $a=c$ is only possible if $3c^2 = 1$ . Thus there are no solutions in this case.

Since the equation $2(a^3 + b^3) = 3ab + 1$ only has the solution $a=b=1$ in integers, putting $x=a$ , $y=2b$ means that the equation $8x^3 - 4 = y(6x - y^2)$ only has the solution $x=1,y=2$ , making the solution $1 + 1 + 2 = \boxed{4}$ .

We can rewrite the equation as a cubic equation (in $y$ ),

$y^3 + y(-6x) + (8x^3 - 4) = 0 .$

Since we're searching for (real) integer solutions, then the cubic discriminant must be non-negative:

$b^2 c^2 -4ac^3 - 4ab^3 d- 27a^2 d^2 + 18abcd \geq 0 .$

In this case $a = 1, b= 0, c = - 6x, d = 8x^3 - 4$ . Upon substituton, we get

$\begin{array} { l c l } -4(1)(-6x)^3 - 27(1^2)(8x^3 -4)^2 &\geq& 0 \\ 32x^3 - (8x^3 - 4)^2 &\geq & 0 \\ -64x^6 + 96x - 16 &\geq & 0 \\ 4x^6 - 6x^3 + 1 &\leq & 0 \\ \frac34 - \frac{\sqrt5}4 \leq x^3 \leq \frac34 + \frac{\sqrt5}4. \\ \end{array}$

The last step above can be found by applying the quadratic formula on the polynomial $5(x^3)^2 - 8(x^3) + 2$ .

Thus the only integer solution of $x$ is 1, and so the value of $y$ is 2. Our answer is $1 + 1 + 2 =\boxed4$ .

Transposing y^3, and adding 12x^2y + 6xy^2 to both sides of the equation, we have:8x^3 + 12x^2y + 6xy^2 + y^3 = 12x^2y + 6 xy^2 + 6xy +4. Then (2x + y)^3 = 6xy(2x + y) = 6xy + 4. Let t = 2x + 5 and r = xy, and add 1 to both sides of the equation, resulting in t^3 + 1 = 6r(t + 1) + 5. Factoring the left side, (t + 1)(t^2 - t + 1) = 6r(t + 1 ) + 5. We conclude that (t +1) must divide 5, whence t +1 = +/1 or +/- 5. The only solution is t + 1 = 5, whence t = 4 and r =2. So the only solution is r = xy = 4, and t = 2x + y = 4, which results in x = 1, y = 2. Since n = 1, x + y + n = 4. Ed Gray