Quadratic, prime numbers

Question

Please be very specific. I wanna ask question no 17 and 18 . In 18 I have an idea but it's not clear to make a perfect solution. Please help.

Answer 1

Hint for 17: it's not hard to calculate $\alpha$ and $\beta$. From that, calculate $P$ and $Q$. The expression you want is $(x-P)(x-Q)$.

Hint for 18: $(x-p) (x-q) = x^2 - (p+q)x + pq$. (And if the coefficient of $x^2$ is not $1$, the sum of coefficients cannot be prime—why?) The sum of coefficients of this expression is $(p-1)(q-1)$. How can $p$, $q$, and $(p-1)(q-1)$ all be prime?

Answer 2

Hint for 17: since $\alpha^2 - 2 \alpha+3=0\,$ we have that $\alpha^2=2 \alpha -3 $. Then:

$$P = \alpha^3-3 \alpha^2+5 \alpha -2 = \alpha(2 \alpha-3)-3(2\alpha-3)+5\alpha-2=2 (2\alpha-3)-4 \alpha+7=1$$

Answer 3

Let $p = 1 + b + c$ and let $\alpha$ and $\beta$ be the roots

Then $-(\alpha + \beta) + \alpha \beta + 1 = p$

If we assume all primes are odd, then

$$p= -(2n + 1 + 2m + 1) + (2m + 1)(2n + 1) + 1 \\= -2(m+ n + 1) + 4mn + 2m + 2n + 1 + 1 = 4mn $$ where $m,n \in \Bbb Z$

Since a prime can't be of form $4mn$ thus $2$ is one of the roots,

$$p = -(2 + \alpha) + 2\alpha + 1 = -(2 + 2m + 1) + (2)(2m + 1) + 1 = -2m - 3 + 4m + 2 + 1 = 2m$$

Hence $p$ is prime and is even thus $m = 1$ which gives $\alpha = 3$ and $\beta = 2$.