$t^3$ 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?

Solutions Collecting From Web of "$t^3$ is never equal to"

You are close, but dont stop with $t^3=at^2+bt$. Since you know what $t^2$ is equal to:
$t^3=a(at+b)+bt=(a^2+b)t+ab$

Now try this for every possiblility.
F.ex. (A): $4=a^2+b$ and $ab=3$ would be possible for $a=1$ and $b=3$.

Since in the all the cases the second term is a prime number either $a$ or $b$ has to be $1$ (since they are both positiv Integers) which makes it easy to check whether it’s possible or not.

Edit: more detailed this gives you a quick solution if it is possible, if it doesn’t give you a quick solution we can try to get a contradiction.

With the method above we see it works for (A), (C) and (D), so we want to check out (B) .

Since obviously $t$ divides $t^3$ we use the form $t^3=8t+5$ and divide it by the $t$. Now we get that $5$ has to be dividable by $t$ (note that normally this is not necessarly true, but here it holds because $8t$ is dividable by $t$).
Because $t$ has to be a positive integer it is now either $1$ or $5$.

Case $t=1$:
Is not possible since $1^3\neq 8+5$

Case $t=5$:
Is not possibe either since $5^3=75\neq45=8*5+5$.

Therefore we get a contradiction, (B) is not possible.

Thank you for pointing this out TonyK.

$t^2 = ax + b$

$t^3 = cx +d$

where $a$, $b$, $c$, $d$ are integers.

See 2 cases:

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

  2. $t^2=at+b$ 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)

$$t^2=at+b\\ t^3=at^2+bt \to t^3=a(t^2)+bt\\t^3=a(at+b)+bt \\t^3=(a^2+b)t+ab$$

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