# Quadratic, prime numbers 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.

• Add your idea to the question – K Split X Apr 17 '17 at 15:25
• Please explain what your idea is. – Henning Makholm Apr 17 '17 at 15:26

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

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

• I think ur calculation is wrong. And can u tell me how do u write all these signs like alpha , mod , under root etc. Thank you for your time – Abhishek Mehta Apr 17 '17 at 15:59
• The calculation is correct as posted, feel free to verify it the hard way by replacing $\alpha$ with $1-i\sqrt{2}$ for example. The way it works is that, given the original equation that $\alpha$ satisfies, you can keep replacing $\alpha^2$ with $2\alpha-3$ until all powers $\ge 2$ are gone. In this case, what's left is just a constant term $1\,$. About formatting, see the MathJax tips. For example, enter $\alpha$ to get $\alpha\,$. – dxiv Apr 17 '17 at 16:03
• Nice answer @dxiv. – A---B Apr 17 '17 at 16:11
• @A---B The same follows from the long division $\alpha^3 - 3\alpha^2+5\alpha-2=(\alpha-1)(\alpha^2-2\alpha+3) + 1$ but it's arguably easier to work out by hand this way. – dxiv Apr 17 '17 at 16:19

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?

• In 17 the roots are imaginary. So that leads us to iota and further it's not going anywhere near to the answer – Abhishek Mehta Apr 17 '17 at 15:53
• " (And if the coefficient of $x^2$ is not $1$, the sum of coefficients cannot be prime—why?) " I don't get the reason why we have to show that $[x^2] = 1$ since every quadratic can be of form $f(x) = (x-p)(x-q)$. – A---B Apr 17 '17 at 17:15
• @A---B it could have the form $2(x-p)(x-q)$ or $3(x-p)(x-q)$ or similar. – Connor Harris Apr 17 '17 at 18:02
• @ConnorHarris Yes you are correct my bad, I confused myself. – A---B Apr 17 '17 at 19:31

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 $\beta = 2$.

• Sorry for the basic question . But have you assumed the equation as "p" and if yes then why 1+b+c – Abhishek Mehta Apr 17 '17 at 16:16
• I understood the whole calculation just wanna know why "p" was taken as 1+b+c – Abhishek Mehta Apr 17 '17 at 16:27
• @AbhishekMehta Sorry my previous comment was not correct, a quadratic can be expressed in the form $f(x) = k(x - \alpha)(x-\beta)$ but if $k \ne 1$ then $k$ is factor of $p$ making $p$ not a prime. – A---B Apr 17 '17 at 19:33
• I got it thank you for your time and helping me. A---B – Abhishek Mehta Apr 18 '17 at 2:52
• @AbhishekMehta You are welcome. You select any one of the answer. – A---B Apr 18 '17 at 13:57