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Old But Gold Polynomials

By giuricrocodilePosted on Tagged

This article is about a couple of problems about polynomials which I regard as highly instructive.

Seems A Demon But It’s Really Not

I first saw this problem in 2018, and I initially thought it was a very hard problem, except it wasn’t. But, at first sight, it could make you feel a bit lost.

Find all the possible values for the expression:

f(x) = \frac{(x-a)(x-b)}{(c-a)(c-b)} + \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)}

where a, b, c are distinct real numbers.

Now, one could try to expand the whole expression and try to finish calculations, but there is a trick which nullifies calculations.

Notice that f(a) = f(b) = f(c) = 1 . That would mean that the quadratic equation f(x)-1 = 0 would have 3 distinct solutions, which is absurd. This implies that

\forall x, f(x) = 1

And, incredible as it may seem, we are done.

Division Between Polynomials

Find the remainder when x^{1959} - 1 is divided by f(x) = (x^2+1)(x^2+x+1) .

Since f(x) has degree four, the remainder has at most degree three.

Hence we can write

x^{1959} - 1 = f(x)q(x) + ax^3+bx^2 + cx + d

Notice that the roots of f(x) are i, -i and the primitive complex third roots of 1, which I will call w, w^2 = -1-w .

If we substitute them in the previous expression we get a system of four equations and four variables because f(x) = 0 for all of them.

\begin{cases} -ai-b+ci+d = -i-1 \\ ai-b-ci+d = i-1 \\ a -(1+w)b +cw + d = 0 \\ a+bw-c(1+w) + d = 0 \end{cases}

Solving this system brings us to the solution

\begin{cases} a=1 \\ b=0 \\ c=0 \\ d=-1 \end{cases}

Hence the remainder is

x^3-1

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