<s>[INST] $t^3$ is never equal to 
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 never equal to:
(A). $4t+3$
(B). $8t+5$
(C). $10t+3$
(D). $6t+5$

We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction?
 [/INST] $t^2 = ax + b$
$t^3 = cx +d$
where $a$, $b$, $c$, $d$ are integers.
See 2 cases: 


*

*$t^2=at+b$ has an integer root $\implies$ t is integer 

*$t^2=at+b$ has no integer root $\implies$ roots are $p+\sqrt{q}$ and
$p−\sqrt{q}$, where $p$, $q$ are rational. You can see that if one
of these roots is a root of $x^3=cx+d$, the second is the root of
$x^3=cx+d$ too, and the third root must be an integer
$\implies$ $t^3-cx-d = 0$ has an integer root
A) $t = -1$
B) No integer root
C) $t= -3$
D) $t= -1$
So, answer is (B)
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