for degree $ p-2 $ there exists the polynomial $ P(x)=x^{p-2} $. now we have to see if exist a polynomial of degree $ p-1 $.

For a similar problem, see the following IMO Shortlist 2010-N4, https://artofproblemsolving.com/community/c6h418642p2362008; and also, the following IMO Shortlist problem, from 1997, which is also solved through very similar moves: https://artofproblemsolving.com/community/c6h49788p315649

Isn't it obvious? Surely Mark Davis.

anonymouslonely wrote: for degree $ p-2 $ there exists the polynomial $ P(x)=x^{p-2} $. now we have to see if exist a polynomial of degree $ p-1 $. If there is no restriction on leading coefficient $T(x) = p x^{p-1} + x^{p-2}$ works nicely ! . But if leading coefficient is not zero mod $p$ , then it's easy there is no polynomial of degree $ p-1 $ which fits into statement . To see how , just note that for all $0<k<p-1$ we have $p \mid S_k = 1^k + 2^k + ...+ (p-1)^k$ . (Demo of proof : Take a primitive root like $g$ and use geometric series formula ) Now if some polynomial $h= \sum_{i=1}^{p-1} a_i x^i$ of degree $p-1$ exist which has non-zero (mod $p$) leading coefficient and preserves all reminder then $0 \equiv S_1=0+1+2+...+(p-1)= \sum_{i=0}^{p-1} h(i) \equiv \sum_{i=0}^{p-1} S_i .a_i\equiv S_{p-1} . a_{p-1} \equiv (p-1)a_{p-1} \not\equiv 0$ Contradiction

mahanmath wrote: anonymouslonely wrote: for degree $ p-2 $ there exists the polynomial $ P(x)=x^{p-2} $. now we have to see if exist a polynomial of degree $ p-1 $. If there is no restriction on leading coefficient $T(x) = p x^{p-1} + x^{p-2}$ works nicely ! . But if leading coefficient is not zero mod $p$ , then it's easy there is no polynomial of degree $ p-1 $ which fits into statement . To see how , just note that for all $0<k<p-1$ we have $p \mid S_k = 1^k + 2^k + ...+ (p-1)^k$ . (Demo of proof : Take a primitive root like $g$ and use geometric series formula ) Now if some polynomial $h= \sum_{i=1}^{p-1} a_i x^i$ of degree $p-1$ exist which has non-zero (mod $p$) leading coefficient and preserves all reminder then $0 \equiv S_1=0+1+2+...+(p-1)= \sum_{i=0}^{p-1} h(i) \equiv \sum_{i=0}^{p-1} S_i .a_i\equiv S_{p-1} . a_{p-1} \equiv (p-1)a_{p-1} \not\equiv 0$ Contradiction What about $(p-a)x^{p-1} + x^{p-2} +a $ $(p-a)x^{p-1} + x^{p-2} +a \equiv\ x^{p-2}\mod p$ so i think this works too and answer should be $p-1$ EDIT:ok it does not work.For $0$ to occur $a$ must be $0$.

It's Pen D 18.