Make It Inconsistent

A nice and original problem!

From the arithmetic progression it follows that $a+c-2b=0$ , so that $x+y+ x^3+y^3 - 2x^2-2y^2=0$ .

Using $y=a-x$ gives a quadratic polynomial in x $(3a-4)x^2+ (4a-3a^2)x + a+a^3-2a^2=0$ with solution

$x=\frac{3a^2-4a \pm \sqrt{(3a^2-4a)^2-4 (3a-4)( a+a^3-2a^2)}}{2( 3a-4)}$ .

We see that the denominator is 0 when $a=\frac{4}{3}$ , but this in itself just means that we do not have a quadratic polynomial. Substituting this value for $a$ back into the polynomial shows that also the linear term vanishes and we are left with the equation $\frac{4}{3}+\frac{64}{27}-\frac{32}{9}=0$ , which is inconsistent indeed because $\frac{4}{27} \neq 0$ . So $a=\boxed{\frac{4}{3}}$ indeed makes it inconsistent.

I still have a question: are x and y to be real numbers? If so, my concern is the discriminant, which has no real solutions when it is negative. It evaluates to $-3a^4+16a^3-28a^2+16a$ , so that for example negative values for $a$ would lead to complex values for $x$ and $y$ . It may be best to always explicitly specify the domain in problems.

Let $x+y=S, xy =P$ . Then $a = S, b = S^2-2P, c = S^3-3SP$ . If they are in an arithmetic progression then $2b=a+c$ , which gives $S^3-3SP+S=2S^2-4P$ . Solving gives us $P=\frac{2S^2-S-S^3}{4-3S}$ , so $S\ =\ \frac{3}{4}$ which gives the answer of 7.

We can find the value of $xy$ using the first and second equation. $xy=\frac{a^{2}-b}{2}$ . Also using the first and third equation we find $xy$ again, $xy=\frac{a^{3}-c}{3a}=\frac{a^{3}-2b+a}{3a}$ .

Equating these two values of $xy$ we will get, $b=\frac{a^{3}-2a}{3a-4}$ , but we don't want any solution of $x$ and $y$ to satisfy the given three equations, so this solution of $b$ should not exist, which is only possible if $a=\frac{4}{3}$ .

Since $a$ , $b$ , and $c$ are in an arithmetic progression, let $b = a + d$ and $c = a + 2d$ . From the first two equations, it can be found that $x$ , $y$ , and $z$ are roots of the equation $2u^2 - 2au + (a^2 - a - d) = 0$ , which after multiplying by $u$ can be rearranged to $u^3 = \frac{1}{2}(2au^2 - (a^2 - a - d)u)$ . Therefore,

$x^3 = \frac{1}{2}(2ax^2 - (a^2 - a - d)x)$

$y^3 = \frac{1}{2}(2ay^2 - (a^2 - a - d)y)$

$z^3 = \frac{1}{2}(2az^2 - (a^2 - a - d)z)$

which combines to

$x^3 + y^3 + z^3 = \frac{1}{2}(2a(x^2 + y^2 + z^2) - (a^2 - a - d)(x + y + z))$

and after substituting the given equations gives

$a + 2d = \frac{1}{2}(2a(a + d) - (a^2 - a - d)a)$

which rearranges to

$d = \frac{a(a - 1)(a - 2)}{3a - 4}$

For $d$ to have a real value, $a \neq \frac{4}{3}$ , and $4 + 3 = \boxed{7}$ .