# Quadratic Prime

We had received some questions on Quadratic equations, But I wasnt able to do one. Here it goes:

Let $a,b$ be natural numbers $a>1$. Also, $p$ is a prime number. If $ax^2+bx+c=p$ for 2 distinct integral values of $x$. Then the number of integral roots of the equation $ax^2+bx+c=2p$ is ? Well I know the answer is $0$, But I am not able to get it properly.

• the condition to have roots... – shooting-squirrel Aug 30 '14 at 4:36
• I tried be way of discriminants, but did not get anything special – Dinesh Aug 30 '14 at 4:37
• the discriminant is strictly bigger than 0, what about now? – shooting-squirrel Aug 30 '14 at 4:40
• Ok, $b^2 > 4a(c-p)$ . and for the second we have $b^2-4a(c-2p)$ – Dinesh Aug 30 '14 at 4:43
• OK, we have 2 real solutions, but not neecessarily 2 $integer$ solutions ... – Dinesh Aug 30 '14 at 4:46

## 2 Answers

Once you are convinced that $a$ divides $b$,$c-p$ and $c-2p$ and that $a=p$, replace $b$ by $p\bar{b}$ and $c-p$ by $p\bar{c}$. Then the equations become (if we divide $p$ out) $x^2+\bar{b}x+\bar{c}$ and $x^2+\bar{b}x+\bar{c}-1$. Then look at the discriminants of both equations: they are $\sqrt{\bar{b}^2-4\bar{c}}$ and $\sqrt{\bar{b}^2-4\bar{c}+4}$. Since the discriminants should be integer, the expressions under the root have to be squares: $\bar{b}^2-4\bar{c}=u^2$ and $\bar{b}^2-4\bar{c}+4 =v^2$ for some integers $u$ and $v$. Then we have that $v^2-u^2=4$. A little calculation shows that $u=0$ and $v=\pm 2$. But $u$ cannot be zero since otherwise the discriminant of the first equation is also zero and both roots of the first equation coincide.

A plan for proving that the latter equation cannot have integer roots, so assuming contrariwise. I left small gaps on purpose.

1. If a polynomial $Ax^2+Bx+C$ with integer coefficients has two integer roots, then we necessarily have $A\mid B$ and $A\mid C$ as $-B/A$ is the sum of the roots and $C/A$ is their product.
2. Show that if the latter equation has at least one integer root, then both of its roots are necessarily integers. From this point on the contrapositive assumption is thus that both equations have two integer roots.
3. Show that this implies that $a=p$.
4. So $$ax^2+bx+(c-p)=p(x-r_1)(x-r_2)$$ and $$ax^2+bx+(c-2p)=p(x-s_1)(x-s_2)$$ for some integers $r_1,r_2,s_1,s_2$. Show that here $r_1+r_2=s_1+s_2$ and $r_1r_2=s_1s_2+1$.
5. So $s_1=r_1+t$ and $s_2=r_2-t$ for some integer $t$. Check that $$1=r_1r_2-s_1s_2=t(r_1-r_2+t).$$
6. Show that the equation in step 5 is impossible unless $r_1=r_2$.
• +1 Great solution. The assumption that $a>1$ is needed to obtain Step 3. – Kim Jong Un Aug 30 '14 at 8:30
• Sir, I understood till point $4$ , but I am a bit confused over point $5$. Can you please add one more line of explanation , please. – Dinesh Aug 30 '14 at 14:41
• @Dinesh: The number $t$ is just the difference $r_1-s_1$. Because $r_1+r_2=s_1+s_2=-b/a$ we have then $r_1-s_1=s_2-r_2$. Plug this in to the equation $$r_1r_2=(c-p)/a=(c-2p)/a+1=s_1s_2+1.$$ – Jyrki Lahtonen Aug 30 '14 at 15:25
• Oh!, Thank you very much, sir. – Dinesh Aug 30 '14 at 15:27