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Consider polynomials which have coefficients

taken modulo 7. So the polynomials

18x12 - 4x5 + x - 1 and 4x12 + 3x5 + x + 6

are equivalent. Note that the exponents are

not taken modulo 7. One such polynomial is:

P[x] = x6 + 4x5 + x4 + 2x3 + 3x2 + 6x + 5.

Can you find a polynomial multiple of P[x]

which is a trinomial and has the smallest

possible degree?

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i know there has to be a logical mathematical way to do this right, not just trial and error? im not that patient

There are better ways to do it than trial and error. I think they all require a bit too much work than you can

do by hand. I think you really need to program a spreadsheet or use a proper programming language to do the

calculations. Once you figure out a way to do it, you'll most likely need a computer to do the grunt work for

you. Of course, someone might figure out a way to do it by hand, but I doubt it.

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In case someone needs it, here's a hint on how to proceed.

Since we are looking for a multiple of P[x], we are actually looking for a polynomial which is 0 modulo P[x]. Whatever the polynomial is, we can subtract as many copies

of P[x] from it without changing its value modulo P[x]. So, P[x] is actually 0 as far as looking at polynomials modulo P[x] goes. Thus, x6 + 4x5 + x4 + 2x3 + 3x2 + 6x + 5 = 0.

Then, x6 = -4x5 - x4 - 2x3 - 3x2 - 6x - 5, or, x6 = 3x5 + 6x4 + 5x3 + 4x2 + x + 2. This allows us to compute all larger powers of x in terms of powers less than 6 as follows:

x7 = x(3x5 + 6x4 + 5x3 + 4x2 + x + 2) = 3x6 + 6x5 + 5x4 + 4x3 + x2 + 2x = 3(3x5 + 6x4 + 5x3 + 4x2 + x + 2) + 6x5 + 5x4 + 4x3 + x2 + 2x. Multiplying and combining mod 7 gives

x7 = x5 + 2x4 + 5x3 + 6x2 + 5x + 6. In the same way, we can compute x8, x9, x10, ...... (as far as we want) in terms of powers of x from x0 to x5. Thus, we can easily

compute the remainder of any power of x when divided by P[x].

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