How do I prove that if $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0≡0 \pmod p$ has more than $n$ solutions, then all $a_i≡0 \pmod p$? Is it possible to solve this using contra positivity, if not what other alternative method is there to approach the problem? I am kind of confused with how to solve it.
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1What have you tried? Alternatively, what is the context behind this problem; where did you find it, and what relevant knowledge do you have? – Christian E. Ramirez Feb 24 '23 at 07:06
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1Is $p$ a prime? Otherwise the statement is false: take $p=6$ and $3x^2-3x$ is always divisible by $6$. – Feb 24 '23 at 07:16
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@StinkingBishop . Or if $p=12$ then $x^2-7x\equiv x^2-7x+12\equiv (x-3)(x-4)\equiv 0\pmod {12}$ has 3 solutions $0,3,4$ mod $12$. – DanielWainfleet Feb 24 '23 at 07:22
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If $p$ is prime then arithmetic on $\Bbb Z_{/p}$ is a $field$. – DanielWainfleet Feb 24 '23 at 07:25
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Iterate the Factor Theorem as in the linked dupes. – Bill Dubuque Feb 24 '23 at 09:16