Maths Olympiad Prep

Library / /92 of 520

Number theory Difficulty 5.6 AIME, harder Prove it

7. Let aa and b=r0b=r_{0} be relatively prime odd positive integers such that
a=r0q1+ϵ1r1r0=r1q2+ϵ2r2rn1=rn1qn1+ϵnrn\begin{array}{l} a=r_{0} q_{1}+\epsilon_{1} r_{1} \\ r_{0}=r_{1} q_{2}+\epsilon_{2} r_{2} \\ \cdot \\ \cdot \\ r_{n-1}=r_{n-1} q_{n-1}+\epsilon_{n} r_{n} \end{array}
where qiq_{i} is a nonnegative even integer, ϵi=±1,ri\epsilon_{i}= \pm 1, r_{i} is a positive integer with ri<ri1r_{i}<r_{i-1}, for i=1,2,,nji=1,2, \ldots, n_{j}, and rn=1r_{n}=1. These equations are obtained by successively using the modified division algorithm given in problem 10 of Section 1.2 .
a) Show that the Jacobi symbol (ab)\left(\frac{a}{b}\right) is given by
(ab)=(1)(r012r112+r112ϵr12++rt112rtr12)\left(\frac{a}{b}\right)=(-1)^{\left(\frac{r_{0}-1}{2} \frac{r_{1}-1}{2}+\frac{r_{1}-1}{2} \frac{\epsilon_{r}-1}{2}+\cdots+\frac{r_{t-1}-1}{2} \cdot \frac{r_{t}^{r}-1}{2}\right)}
b) Show that the Jacobi symbol (ab)\left(\frac{a}{b}\right) is given by
(ab)=(1)T\left(\frac{a}{b}\right)=(-1)^{T}
where TT is the number of integers i,1ini, 1 \leqslant i \leqslant n, with ri1ϵiri3r_{i-1} \equiv \epsilon_{i} r_{i} \equiv 3 (mod4)(\bmod 4).

Solution

None

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.