p^3-4p+9 is a perfect square number find the value of p, for p is positive integer

Question

p^3-4p+9 is a perfect square number  find the value of p, for p is positive integer

anonymous · Answer

@dan815

jdoe0001 · Answer

heheh,  for a perfect square, looks very "cubic"
$\bf p^{\color{brown}{  3?}}-4p+9$

anonymous · Answer

what do you mean ?

anonymous · Answer

yeah, that's p cubic not p square

geerky42 · Answer

Where is $n$ in $p^3-4p+9$?

jdoe0001 · Answer

yeah... the "n" came out of the woodwork, noticed that one

anonymous · Answer

oppss.. sorry that should to find value of p
p^3 -4p + 9 = k^2
i was trying to find value of p, but some method is loss :)

geerky42 · Answer

Funny, this is kind of beyond my knowledge, but by brutal force, I just discovered that p=2 works.

geerky42 · Answer

Because $p^3-4p = (p+2)p(p-2)$, so let p=2 and we have $p-2=0$ so we are left with $9$, which is square number.

geerky42 · Answer

maybe this isn't what you are looking for?

jdoe0001 · Answer

hmmm

can you post a quick screenshot of the material?

geerky42 · Answer

so "brutal force" is ok to you? lol

anonymous · Answer

so, just trials method ?

geerky42 · Answer

to my limited knowledge, yeah, though I'm certain there is better method.

anonymous · Answer

btw, 7 is works too :)

geerky42 · Answer

@amistre64 Do you know any methods?

geerky42 · Answer

Do you just need one value of p that works or all possible values of p?

anonymous · Answer

all values

geerky42 · Answer

then yeah brutal force wouldn't work here, since obviously there are infinity values of p we need to check lol...

geerky42 · Answer

@dan815 @Michele_Laino do you know any methods?

michele_laino · Answer

I try!

michele_laino · Answer

please wait...

michele_laino · Answer

I have found one solution.

michele_laino · Answer

I rewrite your equation as below:
$$\large {x^3} - 4x + 9 = 0$$

geerky42 · Answer

Why set it equal to 0?

michele_laino · Answer

sorry I have misunderstood!

geerky42 · Answer

@texaschic101 @tkhunny @jim_thompson5910

amistre64 · Answer

$$p^3-4p+9=x^2$$
$$p(p^2-4)=x^2-9$$
$$p(p-2)(p+2)=(x-3)(x+3)$$
$$p=\frac{(x-3)(x+3)}{(p-2)(p+2)}$$
$$\frac ab\in Z~\implies~\forall (a,b)\in Z,~and~a\ge b~:~gcd(a,b)=b$$
can we work with this any?

amistre64 · Answer

$$x^2-9~\ge p^2-4$$
$$x^2-5~\ge p^2$$
$$x^2-5~\ge p^2$$
------------------
$$x^2-9= k(p^2-4)$$
$$\frac{x^2-9-4k}{k}= p^2$$
------------------------
$$kx^2-5k\ge x^2-4k-9$$
$$(k-1)x^2-k+9\ge 0$$

can we do anything with this nonesnse?