Find all prime $p$ such that the equation $x^2+px-330p=0$ has exactly one positive integral root.

Question

From the equation the two factors of $-330p$ add up to $p$. But I wasn't able to set up a equation based the conditions given.

How did you type in that exponent of two? – Valkyrie Mar 23 '14 at 19:35 — Valkyrie, Mar 23 '14 at 19:35

score 2 · Answer 1 · answered Mar 23 '14 at 19:09

2

Note that $f(x)=(x-r)(x+p+r)=x^2+px-r(p+r)=0$ is an equation with root $r$.

This is of the form you require iff $r(p+r)=330p$

From this $p|r$ or $p|(p+r)$ - but in the second case we also have $p|r$ or $r=kp$. Hence $$kp(p+kp)=330p; kp(k+1)=330$$

Since then $p|330$ it remains to test $p=2,3,5,11$ to see what works.

I've left some stuff out for you to fill in the gaps, because you didn't give much working or context in the question.

answered Mar 23 '14 at 19:09

Mark Bennet

100,194

Is that the division sign separating p and r? – Valkyrie Mar 23 '14 at 19:42
@Valkyrie $p|r$ means $p$ is a factor of $r$ or equivalently $r=np$ with $n$ an integer. – Mark Bennet Mar 23 '14 at 19:43
You mean 1+r/p is your second case. How do you r/p from it? – Valkyrie Mar 23 '14 at 20:28
@Valkyrie If $p|(p+r)$ then $p+r=np$ and $r=(n-1)p$ is a multiple of $p$. – Mark Bennet Mar 23 '14 at 20:32

DeepSea · Accepted Answer · 2014-03-23T19:16:25.463

1

I find $p = 3$. We have: $x^2 = p(330 - x)$. So $p$ divides $x^2$, and it is a prime so $p$ divides $x$. Write $x = kp$ for some natural number $k$. Then: $(kp)^2 + p(kp) - 330p = 0 $gives: $$pk^2 + pk - 330 = 0 \Rightarrow pk(k+1) = 330$$. But $330 = 3*10*11$. So this happens only when $k = 10$, and $p = 3$. And $x = kp = 10*3 = 30$. But also that $330 = 5*6*11$. So $k = 5$, and $p = 11$ will also work. So $x = kp = 5*11 = 55$ is another answer.

edited Mar 23 '14 at 19:16

answered Mar 23 '14 at 19:09

DeepSea

77,651

Did you not notice $330=5\times 6 \times 11$? – Mark Bennet Mar 23 '14 at 19:11
@MarkBennet: i just went back and complete it – DeepSea Mar 23 '14 at 19:14
Why not p equal to 5 and k equal to 6? – Valkyrie Mar 23 '14 at 19:37
@Valkyrie: you are right. But the main ideas covered. I knew we missed one or two answers. – DeepSea Mar 23 '14 at 19:41
I got it. But...when you were trying to solve my problem, what led you to separate x^2 and substitute it with other variables? – Valkyrie Mar 23 '14 at 21:37
Could you solve the problem based on sum and product of the roots? If so, post it here along with explaination, thanks. I heard about it but did quite get it. – Valkyrie Mar 26 '14 at 22:08

score 1 · Answer 3 · answered Mar 23 '14 at 19:12

The equation has a real positive solution and a real negative solution. If $x$ is an integer solution, then $p$ divides $x$. Let $x=py$. substituting, we find that $p^2y^2+p^2y-330p=0$, and therefore $p(y^2+y)=330$.

It follows that the only candidates for $p$ are $p=2$, $3$, $5$, or $11$. Check them all out.

Find all prime $p$ such that the equation $x^2+px-330p=0$ has exactly one positive integral root.

3 Answers3