Let : $P(x)=x^{3}+ax^{2}+bx+c$ where $(a,b,c)\in Z^3$

Question

Question :

If :

$P(x)=x^3+ax^2+bx+c$ where $(a,b,c)\in Z^3$

And $m,n,k$ root of $P(x)$

such that : $m.n=k$

Then show that : $2P(-1)$ multiple of

$P(1)+P(-1)-2[1+P(0)]$

My try :

We known that :

$m+n+k=-a$

$mn+mk+nk=b$

$m.n.k=-c$

Since $m.n=k$

So : $k^2=-c$

But how I complete this work

Please give me ideas or hints

Answer 1

First let's simplify the things that are required in the proof.

$$P(1) + P(-1) - 2[1+P(0)] = (1+a+b+c)+(-1+a-b+c) -2(1+c)$$

$$=2a-2$$

And,

$$2P(-1) = 2(-1+a-b+c)$$

Now we have to show that $2a-2$ divides $2(-1+a-b+c)$ which is equivalent to saying that $a-1$ divides $b-c$.(Why?)

Now writing out the expressions for $a,b,c$.

$$a = -(m+n+k)$$ $$b=mn+nk+mk$$ $$c = -mnk$$

Therefore,

$$b-c= mn+nk+mk+mnk$$ $$=k+nk+mk+k^2$$

$$=k(1+m+n+k)=k(1-a)$$

Which is what we wanted!