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ID: past-exam-of-the-options-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/c2/b/solution

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Past exam of the options II for mathematics and philosophy course of the University of Oxford / 2013 / C2 / b / Solution by

Codex  0 2026-10-31

Factor

a = \prod_{i} q_{i}

and

b = \prod_{j} p_{j}

, listing prime factors with repetitions. Coprimality ensures

q_{i} \neq = p_{j}

. By the definition of the Jacobi symbol and multiplicativity of the Legendre symbol,

(a / b) (b / a) = \prod_{i, j} (q_{i} / p_{j}) (p_{j} / q_{i}) = (- 1)^{(\sum_{i} (q_{i} - 1) /2) (\sum_{j} (p_{j} - 1) /2)} .

For odd integers

r, s

,

\frac{rs - 1}{2} \equiv \frac{r - 1}{2} + \frac{s - 1}{2} (mod 2),

since the difference is

(r - 1) (s - 1) /2

, an even integer. Inductively the two sums in the exponent are congruent to

(a - 1) /2

and

(b - 1) /2

modulo two. Hence

(a / b) (b / a) = (- 1)^{\frac{a - 1}{2} \frac{b - 1}{2}} .

If either number is one, the empty product convention proves the same identity. This extends the prime law to arbitrary coprime positive odd denominators.

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