longlong123 wrote: Let $a,b,c\in\{0,1,2,\cdots,9\}$.The quadratic equation $ax^2+bx+c=0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number. Wrong. Choose as counter example $(a,b,c)=(0,1,3)$

pco wrote: longlong123 wrote: Let $a,b,c\in\{0,1,2,\cdots,9\}$.The quadratic equation $ax^2+bx+c=0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number. Wrong. Choose as counter example $(a,b,c)=(0,1,3)$ The phrase "QUADRATIC equation $ax^2+bx+c=0$" assume that $a \neq 0$, n'est pas?

nnosipov wrote: pco wrote: longlong123 wrote: Let $a,b,c\in\{0,1,2,\cdots,9\}$.The quadratic equation $ax^2+bx+c=0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number. Wrong. Choose as counter example $(a,b,c)=(0,1,3)$ The phrase "QUADRATIC equation $ax^2+bx+c=0$" assume that $a \neq 0$, n'est pas? I dont know. I think that if $a$ is supposed to be non zero and if this is a serious real olympiad exercice, then $a\ne 0$ should be indicated.

pco wrote: I dont know. I think that if $a$ is supposed to be non zero and if this is a serious real olympiad exercice, then $a\ne 0$ should be indicated. Vous avez raison mais ... cette probleme est assez trivial pour etre "a serious real olympiad exercice". D'accord?

I think $ a\neq 0 $ . Can I do like this?

Click to reveal hidden text

If I made any mistake, please tell me.

We need to prove that there exist a decomposition $ax^2+bx+c=(px+q)(rx+s)$ with $p,q,r,s \in \mathbb{Z}$.

More than that, there exists a decomposition $P(x) = ax^2+bx+c=(px+q)(rx+s)$ with $p,q,r,s \in \mathbb{N}$, since once $P(x)$ has a rational root, the other is also rational, and they both need be non-positive. After that, as in the hidden solution, $P(10) = \overline{abc} = \overline{pq}\cdot \overline{rs}$, where $p,r \neq 0$, and we are done.

It is special case of Cohn Criterion in polynomial. $P(x)=a_nx^n+...+a_1x+a_0$ is irreducible if p-prime is written in base b :$p_b=\overline {a_na_{n-1}...a_1a_0} $