So I’m doing more of a math problem now, so I’m not sure if this is the right place to ask but I thought perhaps somebody could help.

So the problem goes as follows:

Prove that for a polynomial P(x), such that its coefficients are all integers, if there exist at least 4 distinct values of x, such that P(x)=1, there can be no value k for which P(k)=-1.

The proof I found states that if there are some 4 x_i for which this claim holds, namely x_1, x_2, x_3, and x_4 and P(x_i)=1, then for some k we derive:

P(k)-1=(k-x_1)(k-x_2)(k-x_3)(k-x_4)Q(k), such that Q(k) is a polynomial with integer coefficients.

I really don’t understand a thing about this proof. How is this even derived? I’d appreciate your help in deciphering this proof. Thanks.

1 Like

Well, if you’re talking about this step then it’s pretty simple like,

Consider the polynomial F(x)=P(x)-1, now the claim says there are at least 4 such values for which P(x)=1 \implies P(x)-1=0\implies F(x)=0. Hence the inference from the first part is that F(x)=0 has 4 distinct roots \displaystyle x_1,x_2,x_3,x_3. Now since these x_i's are roots hence on factorizing F(x) we’ll get a polynomial which looks something like this,

F(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)Q(x)

Since F(x_i) has to be zero. Now here Q(x) is any polynomial with a degree 4 less than the degree of F. Now since F(x)=P(x)-1

\therefore P(x)-1=(x-x_1)(x-x_2)(x-x_3)(x-x_4)Q(x)

Now I guess we’ll try to prove it by contradiction like, assume for some k , P(k)=-1 so,

P(k)-1=(k-x_1)(k-x_2)(k-x_3)(k-x_4)Q(k) \\
\implies (k-x_1)(k-x_2)(k-x_3)(k-x_4)Q(k)=-2

2 Likes

To add on (assuming we are working in \mathbb{Z}[x], so all x_i's and k are in \mathbb{Z}), this is not possible, because if k is distinct from x_1,x_2,x_3,x_4, then the left hand side can ever be a divisor of -2 (this can be done by casework as to which side of k the x_i's lie).

2 Likes

I completely forgot about existence of roots of polynomials. Thanks for the proof though. It’s all clear now.

1 Like

If I’m not mistaken then ig only coefficients are in Z not the roots

My bad, x can take only integer values so it’s also the roots. Sorry for omitting that detail.

1 Like

Correct. My point holds only if x_i's are assumed to be integers.

Yes then next part is trivial, as we can subdivide -2 as a product of at max three distinct integers 2,1,-1 hence any integer more than 3 can’t be the number of roots.

2 Likes