Write down the most general polynomial of degree 4 that leaves a remainder of 1 when divided by any of
\(x-1\,\), \(x-2\,\), \(x-3\,\) or \(x-4\,\).
The polynomial \(\P(x)\) has degree \(N\), where \(N\ge1\,\),
and satisfies
\[
\P(1) = \P(2) = \cdots = \P(N) =1\,.
\]
Show that \(\P(N+1) \ne 1\,\).
Given that \(\P(N+1)= 2\,\), find \(\P(N+r)\) where \(r\) is a positive integer. Find a positive integer \(r\), independent of \(N,\) such that \(\P(N+r) = N+r\,\).
The polynomial \({\rm S}(x)\) has degree 4. It has integer coefficients and the coefficient of \(x^4\) is 1. It satisfies
\[
{\rm S}(a) =
{\rm S}(b) =
{\rm S}(c) =
{\rm S}(d) = 2001\,,
\]
where \(a\), \(b\), \(c\) and \(d\) are distinct (not necessarily positive)
integers.
Show that there is no integer \(e\) such that \({\rm S}(e) = 2018\,\).
Find the number of ways the (distinct) integers \(a\), \(b\), \(c\) and \(d\) can be chosen such that \({\rm S}(0) = 2017\) and \(a < b< c< d\,.\)
Solution:
\(p(x) = C(x-1)(x-2)(x-3)(x-4)+1\)
Suppose \(P(N+1) = 1\) them we could consider \(f(x) = P(x) - 1\) to be a polynomial of degree \(N\) with at least \(N+1\) roots, which would be a contradiction. Therefore \(P(N+1) \neq 1\).
Since \(P(x) = C(x-1)(x-2)\cdots(x-N) + 1\) and \(P(N+1) = 2\) we must have \(C \cdot N! + 1 = 2 \Rightarrow C = \frac{1}{N!}\), hence \(P(x) = \binom{x-1}{N} + 1\) ie \(P(N+r) = \binom{N+r-1}{N}+1\) so \(P(N+2) = \binom{N+1}{N} +1= N+2\), so we can take \(r=2\).
Suppose consider \(p(x) = S(x) - 2001\), then \(p(x)\) has roots \(a,b,c,d\) and suppose we can find \(e\) such that \(p(e) = 17\) then we must have \((e-a)(e-b)(e-c)(e-d) = 17\) but the only possible factors of \(17\) are \(-17,-1,1,17\) and we cannot have all \(4\) of them. Hence this is not possible.
Now we have \(abcd = 16\), so we can have factors \(-16,-8,-4, -2, -1, 1, 2, 4,8,16\) (and we need to have \(4\) of them).
If we have \(0\) negatives, the smallest product is \(1 \cdot 2 \cdot 4 \cdot 8 > 16\)
If we have \(2\) negatives we must have \(1\) and \(-1\) (otherwise we have the same problem of being too large. So \(\{-1,1,-2,8\},\{-1,1,2,-8\},\{-1,1,-4,4\},\)
If we have \(4\) negatives that's the same issue as with \(0\) negatives.