# Proof that polynomial with given rules is not possible

Prove that there is no polynomial $$P(x)$$ with whole number coefficients for which: $$P(7)=5\\P(15)=9$$

So what I know, a polynomial is defined as:

$$P(x)=a_0+a_1x+a_2x^2+...+a_nx^n$$

and we need $$a_0,a_1,...,a_n\in\Bbb{Z}$$

so first I wrote down:

$$P(7):a_0+7a_1+7^2a_2+...+7^na_n=5$$

thought of simplyfing this somehow to make it usable in any way

$$7(a_1+7a_2+...+7^{n-1}a_n)+a_0=5\\7(a_1+7(a_2+7(a_3+...+7a_n)))+a_0=5$$

I thought now of maybe trying to move something to the other side

$$a_1+7(a_2+...+7a_n)={5-a_0\over7}$$

I could go and do this forever now and I'm not sure how to connect that with proving that $$P(x)$$ with the given rules doesn't exist. Stuck here.

• I think there is no polynomial $P(x)$ with whole number coefficients for which $P(7)=5$.(except the constant polynomial)
– math
Commented Jan 16, 2020 at 19:23
• does whole number mean element of $\{0,1,2,3,\dots\}$? Commented Jan 16, 2020 at 19:29
• They meant $\Bbb{Z}$ I'm pretty sure Commented Jan 16, 2020 at 19:34
• well then what about $x^2-7x+5,$ @math ? Commented Jan 16, 2020 at 19:36
• sorry, I thought whole numbers means non-negative [email protected]
– math
Commented Jan 16, 2020 at 19:38

Hint:

If $$p\in \mathbb{Z}[x]$$ then for every integers $$x,a$$ we have $$x-a\mid p(x)-p(a)$$

This fact is easy to prove. If we divide $$p(x)$$ with $$x-a$$ we get remainder to be a constant say $$c$$ so we have $$p(x) = k(x)(x-a)+c\;\;\;\;(*)$$ Now, what is $$c$$? Set in $$(*)$$ $$x=a$$ and we get $$c= p(a)$$ and we are done.

• Oh got it now! Took me a while to understand what you hinted at. Thank you! Commented Jan 16, 2020 at 19:32