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Suppose $f$ is a polynomial over $GF(17)$ of degree at most two. Find $f$ given that $f(3)=5, f(2)=6$ and $f(8)=1$.

I have tried trial and error but had no luck, I was wondering if there was a better method? I know $GF(17)$ is a cyclic group of order $17^2-1=288$, don't know if I can use this to help me?

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    FYI:The multiplicative group of $GF(17)$ is cyclic of order $17-1=16$. Anyway, why don't you try [Lagrange interpolation polynomial](https://en.wikipedia.org/wiki/Lagrange_polynomial). All you need to do is to plug in the numbers. – 2017-02-18
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    You could also write $y=ax^2+bx+c$ and get three data points $(x_i,y_i)$, allowing you to solve for $a$, $b$, and $c$. But @JyrkiLahtonen’s suggestion is quickest. – 2017-02-18

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$$ f(x) = \frac{1}{30} (x-3)(x-2) + (x-3)(x-8) - (x-2)(x-8) $$

I would explain the method I used to come up with this, but I believe it is fairly obvious...

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    I realise now I have to use the Lagrange interpolation formula, I am plugging in the numbers but am confused when I plug my $x_i$ into the denominator of $f_i(x)$ as I get 0's, i.e ($x_0 - x_0$) – 2017-02-18
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    @harry55: Read the formula more carefully. That zero factor is explicitly left out! – 2017-02-18
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    I have got to the answer now, so the fact it is over $GF(17)$ makes no difference to the process? Say it was over $GF(5)$, then it would still be the same? – 2017-02-18
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    Over $GF(5)$ we have $8 = 3$, so it's not possible with the same values. – 2017-02-18
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    I have found out from my lecturer that this answer is wrong as it does not contain elements in $GF(17)$. Could anybody provide me with the right answer? – 2017-03-02
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    @harry55 I am not sure what that means. If he takes issue with how $ 1/30 $ shows up here, well, $ 30 $ is invertible in $ \mathbb F_{17} $ with inverse $ 4 $... – 2017-03-02