Which of the following value(s) of $t^3$ are not possible? Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is not equal to -
1) $4t+3$
2) $8t+5$
3) $10t+3$
4) $6t+5$
This question is from a prestigious Indian Scholarship exam, and I'm solving it for fun. But I can't get at the bottom of the problem. I'm also a bit shaky on which concept to use in this problem.
My try on this question 
$t^2 =at +b$
$t = \sqrt(at+b)$   
For $t \in R$
$at+b \ge 0$
Since $a$ and $b$ are positive integers 
$at \ge -b$  and
$t \ge \frac{-b}{a}$
I also tried differentiating the function    $t^2 = at + b$ to get
$\frac{d}{dt}(t^2) = a\frac{d}{dt}(t) + 0$
$2t = a$
Alas it leads nowhere. But
$\frac{a}{2} \ge \frac{-b}{a}$
Since $a$ and $2$ are positive integers
$a^2 \ge -2b$
What has to done next? Please help 
 A: Hint:$$t^3 = t^2\cdot t \\= (at+b)\cdot t \\= at^2 + bt \\= a(at+b) + bt \\= (a^2 + b)t + ab$$
A: Assume that $t^3$ does equal $ut+v$ (where $u=4$, $v=3$ for the first part, etc.). 
Then $t$ is a root of the cubic polynomial
$$f(X)=X^3-uX-v \in\Bbb Z[X]$$
as well as of the quadratic polynomial
$$g(X)=X^2-aX-b \in\Bbb Z[X].$$
But then $t$ is also a root of 
$$f(X)-Xg(X) =aX^2+(b-u)X-v$$
and also of
$$h(X)=(f(X)-Xg(X))-ag(X)=(a^2+b-u)X+ab-v. $$
There are two possibilities:


*

*If $a^2+b=u$ then we conclude $ab=v$. Two of the problem parts allow you to find $a,b$ with these properties. As the corresponding $g(X)$ has two real roots (the discriminant $a^2+4b$ is certainly positive), in these cases it is possible that $t^3=ut+v$.

*If $a^2+b\ne u$, $h(X)$ has a unique root $t$ and it is rational. By the rational root theorem, $t$ must in fact be an integer and among the signed divisors of $v$. Conveniently, $v$ is prime in all problem parts, you you need only check for $t\in\{v,1,-1,-v\}$ whether or not $t^3=ut+v$. That is, check if $v^2=u+1$ or $u=0$ or $v=u-1$ or $v^2=u-1$.
