<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] $$t^2=at+b\\ t^3=at^2+bt \to t^3=a(t^2)+bt\\t^3=a(at+b)+bt \\t^3=(a^2+b)t+ab$$
$$t^3=(a^2+b)t+ab=4t+3 \to ab=3 \to \\a=1 ,b=3 \to t^3=(1^2+3)t+1.3 \\$$
$$t^3=(a^2+b)t+ab=10t+3 \to ab=3 \to \\a=3 ,b=1 \to t^3=(3^2+1)t+1.3 \\$$
$$t^3=(a^2+b)t+ab=6t+5 \to ab=5 \to \\a=1 ,b=5 \to t^3=(1^2+5)t+1.5 \\$$
so answer is :"B"
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