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If it's infinite, is it countable or uncountable infinite?

I am a newbie to this topic... I don't know what modular arithmetic for polynomials means. Can someone please give me a link where I can learn?

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Well, it means the coefficient arithmetic is modular arithmetic. I.e. the polynomials have coefficients in the field $\Bbb{Z}/7\Bbb{Z}$. –  user2468 Aug 26 '12 at 3:26
Thanks. This kind of helps me understand. Although I don't really know what field means and the notation z/7z is new to me. –  darksky Aug 26 '12 at 3:35

3 Answers 3

up vote 1 down vote accepted

There are $8$ coefficients to be determined. The lead coefficient cannot be $0$. So the number is $(6)(7^7)$.

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Thanks. So I presume the coefficients must be non-negative integers less than 7? –  darksky Aug 26 '12 at 3:20
Without loss of generality we can take them to be as you described. In principle they are abstract objects $[0], [1], \dots, [6]$. –  André Nicolas Aug 26 '12 at 3:27

You only have 6 or 7 (why not always 7?) choices for the coefficients. Thus the number is finite, and you should be able to figure it out for yourself....

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-1. To someone who doesn't understand the topic and doesn't know what modular arithmetic, how is this answer any use at all? If "you should be able to figure it out for yourself" were true, would the OP have bothered posting? –  user22805 Aug 26 '12 at 3:46
Well, I added the second paragraph of my post in an edit after this answer was posted. Mostly like he didn't know –  darksky Aug 26 '12 at 3:53
@DavidWallace When I answered the post was only consisting of the first question. And since it looked like a simple, potential homework question, I didn't provide a full answer. I guess in the future I should try to guess potential future edits ;) –  N. S. Aug 26 '12 at 12:27

All such polynomials look like: $$ a_1 x^0 + a_2 x^1 + \cdots + a_n x^{n-1} + a_{n+1} x^{n}$$ where $a_i \in \{ \color{blue}{0}, 1, \ldots, n-1 \}$ for $1 \le i \le n$ and $a_{n+1} \in \{1, 2, \dots, n-1 \}.$

So there are $$\underbrace{n \times n \times \cdots \times n}_{n\text{ times}} \times (n-1) = n^n \times (n-1)$$ such polynomials.

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