Question

Let $t$ be 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. $6t+5$

Answer 1

The correct option is B $8t+5$

$t^2=at+b{t}^3=a{t}^2+bt=a(at+b)+bt=(a^2+b)t+ab;a,b\in Z^{+}\left(a\right)a^2+b=4ab=3\text{possible for}\left(\right(a,b)=(1,3\left)\right)\left(b\right)a^2+b=8ab=5\text{not possible}\left(c\right)a^2+b=10ab=3\text{possible for}\left(\right(a,b)=(3,1\left)\right)\left(d\right)a^2+b=6ab=5\text{possible for}\left(\right(a,b)=(1,5\left)\right)$