Bourbaki

Exercise 24, § 1

Algebra, VII, p. 53

a) Let n=∏kpkvk>0n = \prod_k p_k^{v_k} > 0 be a rational integer expressed in terms of its prime factors. Show that the sum of the positive divisors of n is given by the formula

σ(n)=∏kpkvk+1−1pk−1. \sigma(n) = \prod_k \frac{p_k^{v_k+1} - 1}{p_k - 1}.
b) We say that n is perfect if 2n=σ(n)2n = \sigma(n). Show that an even number n is perfect if and only if it has the form 2h(2h+1−1)2^h(2^{h+1}-1), where h≥1h \geq 1 and 2h+1−12^{h+1}-1 is prime. (If n has the form 2mq2^m q, where m≥1m \geq 1 and q is odd, show that q is divisible by 2m+1−12^{m+1}-1, and deduce that, if q were not prime, then we would have σ(n)>2n\sigma(n) > 2n).

Solution

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Transcribed from the printed volume. Not translated.