# I Primes and Polynomials

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1. Dec 29, 2017

### Staff: Mentor

It is your problem, I think you should contribute something as well. Finding the right approach is the interesting part of mathematics.

It could help to look at other easier cases first, e.g. third degree polynomials.

2. Dec 29, 2017

### JimBob81345

I think I should use a different approach than the one used previously.

Last edited by a moderator: Dec 29, 2017
3. Dec 29, 2017

### JimBob81345

mfb, I believe I got it. Can you please check over this solution?
for deg(h) = 1, we have h(x) = k(x-11) which gives k for h(12) (k must be an integer). Since x-11 is an integer for all integer x, k(x-11) must be divisible by k, thus, k(x-11) must be an integer.
let y_i denote the i-th number in A.
Assume h(x) is an integer for some x=k.
This is what we want to prove. If k(x-11)(x-12)...(x-y_i) is an integer, then k(x-11)(x-12)...(x-y_i)(x-y_(i+1)) is an integer. (proof by indiction).
(x-y_(i+1)) is an integer because x and y_(i+1) are integers, and if you multiply an integer by an integer you get an integer. So our proof is done for all cases.

4. Dec 29, 2017

### Staff: Mentor

You set x=k?
The k might change depending on i. Induction might work, but it needs more thought to make a solid proof.

There is another thing that I didn't prove properly - I'm highly confident it is true, but a proof would be better: Show that g(x) is always an integer for all integer x. This is clear if f is a parabola, and it is easy to prove if we take n subsequent integers where h(x) is zero, but the general case needs some more thought.

Where does the problem come from?

5. Dec 30, 2017

### JimBob81345

Sorry x is not equal to k
I am delving into research mathematics and this is a problem I proposed to myself

6. Dec 31, 2017

### JimBob81345

Can a be an irrational number in f(x) = a(x-11)(x-12)...(x-y_n)?

7. Dec 31, 2017

### Staff: Mentor

That would make the function irrational at all integers (at all rational numbers, even) where it is not zero.