Finding the number of polynomial when $f(a) , f(b)$ is given

Question

Find the number of polynomials given by $f(x)$ , with integral coefficients. Such that $$f(7) =11; f(11)=13$$

Please suggest the solution of this..

Thanks ..

It won't help you find the solution but you can notice if you have one solutions $f$, then all the polynomials $f_n(x)=f(x)+(x-7)^n(x-11)^n, n\in \Bbb N^*$ are solutions. So you can either have $0$ solutions or infinitely many. — xavierm02, Apr 28 '13 at 11:47

clark · Answer 1 · 2013-04-28T11:54:26.297

6

hint Use the fact that
$$m-n | f(m)-f(n)$$

To see this use the identity $m^k-n^{k}= (m-n)(m^{k-1}+m^{k-2}n+\cdots+n^{k-1})$

edited Apr 28 '13 at 11:54

answered Apr 28 '13 at 11:43

clark

that means, (13-11)|11-7 = 2|4 Could u please some more detail about the concept.. Thanks.. – Sachin Apr 28 '13 at 11:49
No, it's the other way around. $4 \mid 2$ which is false. – xavierm02 Apr 28 '13 at 11:49

score 3 · Accepted Answer · answered Apr 28 '13 at 11:57

$f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots a_0$

$f(7)=a_n7^n+a_{n-1}7^{n-1}+\dots a_0=11$

$f(11)=a_n11^n+a_{n-1}11^{n-1}+\dots a_0=13$

$f(11)-f(7)=a_n(11^n-7^n) +a_{n-1}(11^{n-1}-7^{n-1})+ \dots a_1\cdot4=2$

Now you know that $a-b|a^n-b^n$, from here.

$4|f(11)-f(7) \implies f(11)-f(7)=4k$

But $\dots 4 \not| 2$, conclude.

2 Answers2