4

If p is a prime and both roots of $x^{2}+px−444p=0$ are integers, what is $p$

I got that for the roots to be integers the discriminant must be a perfect square. Thus, $p(p + 1776)$ must be a perfect square.

However, at this point I am stuck and do not know how to proceed other than trying values for $p$ and checking what works.

1110101001
  • 4,188

4 Answers4

9

HINT:

We have $\displaystyle q^2=p(p+1776)$ where $q$ is some integer

As $p$ is prime, $\displaystyle p$ must divde $q,q=pr$(say)

$\displaystyle\implies 1776p=p^2(r^2-1)\iff \frac{1776}p=r^2-1$ which is an integer

Check for the prime factors of $1776$

lab bhattacharjee
  • 274,582
  • I don't follow how you got that 1776p=p^2 * (r^2−1) – 1110101001 Feb 16 '14 at 06:55
  • Why does p|x^2 mean p|x only when p is prime? How come when p is not prime this does not hold true? – 1110101001 Feb 16 '14 at 07:03
  • @user2612743, for the 1st Question, put $q=pr$ $$\implies (rp)^2=p^2+1776p\iff p^2(r^2-1)=1776p$$$ – lab bhattacharjee Feb 16 '14 at 15:05
  • @user2612743, for the second if the highest power of prime $p$ that divides $x$ is $h,$ then the highest power of prime $p$ that divides $x$ is $2h$ So, if $p|x^2, 2h\ge1\implies h\ge1$ .But this may not hold if $p$ is not prime. For example, if $8|x^2,$ we can not say $8|x$ – lab bhattacharjee Feb 16 '14 at 15:20
  • @lab That prime $,p\mid x^2,\Rightarrow,p\mid x,$ is an immediate consequence of the uniqueness of prime factorizations, a property which is usually considered a bit more fundamental than the property you mention (that view comes more to the fore in valuation theory). – Bill Dubuque Feb 16 '14 at 16:01
3

Factor it as $(x-a)(x-b)$. Then, $ab=-444p$, so one of $a,b$ is a multiple of $p$. But since $a+b =-p$, they are both multiples of $p$.So $p$ is a divisor of $444$(because $-444p$ would be a multiple of $p^2$)

chubakueno
  • 5,623
3

$ x^2\!+px = 444p\,\Rightarrow \,p\mid x^2\,\Rightarrow\,p\mid x\,\Rightarrow\ p^2\mid 444p\,\Rightarrow\, p\mid 444\,\Rightarrow\, X^2\! + X = 444/p,\ X = x/p,\, $ which has roots $\,m,n\,$ with $\ \color{#C00}{m+n} = -1,\,\ mn = -444/p = \color{#c00}{- 4\cdot 3}\cdot 37/p,\ $ so $\ \color{#c00}{m,n},p \, =\, \ldots$

Bill Dubuque
  • 272,048
  • How do you know/prove that if p divides x^2 then p also divides x? – 1110101001 Feb 16 '14 at 06:59
  • In other words, why does p|x^2 mean p|x only when p is prime? How come when p is not prime this does not hold true? – 1110101001 Feb 16 '14 at 07:02
  • @user2612743 That's a consequence of the Fundamental Theorem of Arithmetic, in particular, uniqueness of prime factorizations (or equivalent properties such as Euclid's Lemma). – Bill Dubuque Feb 16 '14 at 15:08
2

By Eisenstein's criterion, the polynomial $x^2 + px - 444p$ is irreducible unless $p^2|444p$. If both roots of the polynomial are integers, it is not irreducible and hence $$p^2|444p \;\;\iff\;\; p | 444 \;\;\implies\;\; p = 2, 3 \text{ or } 37$$ So we have only 3 primes to check.

achille hui
  • 122,701
  • That would have been my choice too were it not for the algebra-precalc tag (rarely is Eisenstein known at that level). But that's not much of a loss here, since one can make the same deduction elementarily by comparing powers of $p$ in prime factorizations, e.g. as in my answer, or chubakuenko's. But +1 anyhow! – Bill Dubuque Feb 16 '14 at 15:44