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... Aug 30, 2014 at 4:36
• I tried be way of discriminants, but did not get anything special Aug 30, 2014 at 4:37
• the discriminant is strictly bigger than 0, what about now? Aug 30, 2014 at 4:40
• Ok, $b^2 > 4a(c-p)$ . and for the second we have $b^2-4a(c-2p)$ Aug 30, 2014 at 4:43
• OK, we have 2 real solutions, but not neecessarily 2 $integer$ solutions ... Aug 30, 2014 at 4:46

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. Aug 30, 2014 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. Aug 30, 2014 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 $$r_1r_2=(c-p)/a=(c-2p)/a+1=s_1s_2+1.$$ Aug 30, 2014 at 15:25
• Oh!, Thank you very much, sir. Aug 30, 2014 at 15:27

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.