QUESTIONS AND PROBLEMS 1837.4. Draw dot figures for the first five heptagonal and octagonal numbers. What
kind of figure would you need for nonagonal numbers?
7.5. Prove the formulas given in the caption of Fig. 1 for Tn, Sn, Pn, and Hn. Then
prove that Sn = Tn + Tn.u Pn = Sn + T„_x = Tn + 2T„i, Hn = Pn + Tn. 1 =
Tn + 3Ôç÷. If Pk,n is the nth fc-gonal number, give a general formula for Pk<n in
terms of k and n.
7.6. Prove that the Pythagorean procedure always produces a perfect number.
That is, if ñ = 2 n — 1 is prime, then TV = 2n_1p is perfect. This theorem is not
difficult to prove nowadays, since the "parts" (proper divisors) of TV are easy to list
and sum.
7.7. Let Nn be the nth perfect number, so that N\ = 6, TV 2 = 28, TV 3 = 496,
N4 = 8128. Assuming that all perfect numbers are given by the Pythagorean
formula, that is, they are of the form 2"~^1 (2" - 1) when 2" - 1 is a prime, prove
that /V„+i > 16iVn if ç > 1. Conclude that there cannot be more than one fc-digit
perfect number for each k.
7.8. (V. A. Lebesgue's proof of Eider's theorem on even perfect numbers) Suppose
that the perfect number TV has the prime factorization TV = 2Qp^^1 · · ·ñ£\ where
pi,... ,Pk are distinct odd primes and a, n\,..., n*, are nonnegative integers. Since
Í is perfect, the sum of all its divisors is 2N. This means that
2 Ù+1ÑÃ ·'-P? = (1 + 2 + ·· · + 2°)(1 +pi + ••• +pT)-• "(1 + Pk + • • - + pnkk)
= (2á+' -i)(i+p1 + ...+p»-)...(i+pfc + ... + P2*).Rewrite this equation as follows:
(2
Q+1
-i)Pr ••·ÑÃ+ÑÉ
1
···ÑÃ =
= (2°+] -À)(À+ñ 1 + ...+ñ»·)...(À+ñ<! + ...+ñ^),ÑÃ-ÑÃ + ^+Ã^ =0+ÑÉ+···+Ñ?,)···(É+Ñ* + ·-·+Ñ^)·
Since the second term on the left must be an integer, it follows that 2 á+1 — 1
must divide p"^1 · • ·ñ£*. This is not a significant statement if a = 0 (TV is an odd
number). But if TV is even, so that a > 0, it implies that 2°+1 - 1 = ñøé · ·-p™k
for integers m\ < n\,..., ôç* < ç*,, not all zero. Thus, the left-hand side consists
of the two distinct terms p"^1 · • • p£k + p\x · · • prkk. It follows that the right-hand side
must also be equal to this sum. Now it is obvious that the right-hand side contains
these two terms. That means the sum of the remaining terms on the right-hand
side must be zero. But since the coefficients of all these terms are positive, there
con be only two terms on the right. Since the right-hand side obviously contains
(nj + l)(n2 + 1) · · • (nk + 1) terms, we get the equation2 = (n 1 + l)(n 2 + l)---(nfc + l).
Deduce from this equation that TV must be of the form 2n_1(2" -1) and that 2n - 1
is prime.7.9. Generalize Diophantus' solution to the problem of finding a second represen-
tation of a number as the sum of two squares, using his example of 13 = 2^2 + 3^2
and letting one of the numbers be (ò + 3)^2 and the other (Á;ò - 2)^2.