Show that $q^4+2pq^2 +p^2 = 2pq -(pq)^2 -1$ becomes $p^3+q^3+3pq-1=0$.

Question

I know that these two are exactly the same equation but I can't seem to prove it.

You are also given that $p+q=1$.

This is a follow up from a similar question.

@BillDubuque Hi there, 'Becomes'in this case means 'equal' I wanted to avoid using that word since there are already equal signs in both equations. I know that these two are definitely the same equation. I just cannot seem to prove it. — BLAZE, Aug 11 '14 at 15:46

score 4 · Answer 1 · edited Apr 13 '17 at 12:21

4

HINT:

$$(q^2+p)^2+(pq-1)^2=0$$

Also, $$a^3+b^3+c^3-3abc=(a+b+c)(\sum a^2-bc)$$

Here $a=p,b=q,c=-1$

edited Apr 13 '17 at 12:21

Community

answered Aug 11 '14 at 09:59

lab bhattacharjee

I'm sorry i'm still a little lost on this one, would you care to complete the proof for me even if this is not in my best interest? – BLAZE Aug 11 '14 at 11:07
@user144533, Where is your confusion – lab bhattacharjee Aug 11 '14 at 11:32
my confusion is how $$a^3+b^3+c^3-3abc=(a+b+c)(\sum a^2-bc)$$ can help me obtain $p^3+q^3+3pq-1=0$. – BLAZE Aug 11 '14 at 12:44
1

@user144533, We have $p+q+(-1)=0\implies p^3+q^3+(-1)^3=3pq(-1)$ – lab bhattacharjee Aug 11 '14 at 12:46
Yes. But what does that have to do with $$(q^2+p)^2+(pq-1)^2=0?$$ – BLAZE Aug 11 '14 at 15:57
@user144533, No – lab bhattacharjee Aug 11 '14 at 16:10
Youv'e said:- $$(q^2+p)^2+(pq-1)^2=0$$ Also, $$a^3+b^3+c^3-3abc=(a+b+c)(\sum a^2-bc)$$ But where is the link between $$(q^2+p)^2+(pq-1)^2=0$$ and $$a^3+b^3+c^3-3abc=(a+b+c)(\sum a^2-bc)?$$ – BLAZE Aug 11 '14 at 22:22

score 4 · Accepted Answer · answered Aug 11 '14 at 11:37

4

As lab said.

$$(q^2+p)^2+(pq-1)^2=0$$

Since $p$ and $q$ are real numbers, then LHS is sum of two squares, wich are non-negative, therefore:

$$\begin{cases} q^2 + p = 0 & \Longrightarrow & p = -q^2 \\ pq - 1 = 0 & \Longrightarrow & -q^3-1 = 0 \end{cases}$$

Therefore $q = -1$ what means that $p = -1$, wich contradicts that $p+q = 1$.

But the second equation holds:

$$p^3+q^3+3pq-1= (-1)^3 + (-1)^3 + 3(-1)(-1) -1 = 0$$

answered Aug 11 '14 at 11:37

Darth Geek

$$\begin{cases} pq - 1 = 0 & \Longrightarrow & -q^3-1 = 0 \end{cases}$$ Should this not imply that $pq=1?$ and this line: $$p^3+q^3+3pq-1= (-1)^3 + (-1)^3 + 3(-1)(-1) -1 = 0$$ is more of a verification than proof, what i'm looking for is algebraic proof that $q^4+2pq^2 +p^2 = 2pq -(pq)^2 -1$ is the same equation as $p^3+q^3+3pq-1=0$ – BLAZE Aug 12 '14 at 01:09
@user144533 The implication that $-q^3-1=0$ comes from substituting $p=-q^2$ from the previous line into $pq-1=0$. – Théophile Aug 12 '14 at 01:29
1

@user144533 Well, they are not the same equation. The second one is equivalent to:
$$\frac{1}{3}·(p+q-1)\left((p-q)^2 + (q+1)^2 + (p+1)^2\right)=0$$ Wich is true for $p = q = -1$ or for $p+q = 1$. Whereas the first one, as we've seen, only holds for $p = q = -1$. This is because $p,q\in \mathbb{R}$. Otherwise we'd have more solutions, such as: $$p = \frac{1\pm \sqrt{3}i}{2} \hspace{1cm} q = \frac{1\mp \sqrt{3}i}{2}$$ wich would satisfy $p+q = 1$
– Darth Geek Aug 12 '14 at 07:12
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That is, allowing $p$ and $q$ to be complex numbers. – Darth Geek Aug 12 '14 at 07:18
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Also, note that if we substitute $p = 1-q$ on both equations we get to $(q^2-q+1)^2 + (-q^2+q-1)^2 = 0$ on the first one, but on the second one we get to $0 = 0$ because all the terms cancel out. – Darth Geek Aug 12 '14 at 07:23

Bill Dubuque · Answer 3 · 2014-08-12T01:25:12.547

1

Write the equations as $\,f(p,q) = 0\,$ and $\,g(p,q) = 0,\,$ respectively.

Then $\ {\rm mod}\ p\!+\!q\!-\!1\!:\,\ f \equiv 2(q^4\!-q^2\!+1)\,$ and $\, g\equiv 0$.

So if $\,p\!+\!q=1\,$ then $\,f = g \iff q^4\!-q^2\!+1=0\iff q^6=-1,\ q^2\ne -1$

edited Aug 12 '14 at 01:25

answered Aug 11 '14 at 13:03

Bill Dubuque

Sorry, I'm still a little lost. How does this help me show that $q^4+2pq^2 +p^2 = 2pq -(pq)^2 -1$ is the same equation as $p^3+q^3+3pq-1=0$ – BLAZE Aug 12 '14 at 01:20
1

@user144533 $\ $ If $,p+q=1,$ then $,f=g\iff q^4-q^2+1 = 0\ \ $ – Bill Dubuque Aug 12 '14 at 01:24

3 Answers3