QUESTIONS AND PROBLEMS 219
letting m = r^3 - (p^3 + q^3 ) and ç = 2pqr, verify from the original equation that
m^3 = 3n^3 , which by Problem 8.5 is impossible if m and ç are nonzero. Hence
ç = 0, which means that ñ = 0 or q = 0 or r = 0, that is, at least one of χ and y
equals 0. Conclude that no such positive integers x, y, and æ can exist.
8.7. Verify that
275 + 84^5 + 110^5 + 133^5 = 144^5.
[See L. J. Lander and T. R. Parkin, "Counterexample to Euler's conjecture on sums
of like powers," Bulletin of the American Mathematical Society, 72 (1966), p. 1079.
Smaller counterexamples to this conjecture have been discovered more recently.]
8.8. Prove Fermat's little theorem by induction on a. [Hint: The theorem can be
restated as the assertion that ñ divides ap — a for every positive integer a. Use the
binomial theorem to show that (á + 1)ñ-(á+1) = ôçñ + ap - a for some integer
m.}
8.9. Verify the law of quadratic reciprocity for the primes 17 and 23 and for 67
and 71.
8.10. Show that the factorization of numbers of the form m + n\/^3 is not unique
by finding two different factorizations of 4. Is factorization unique for numbers of
the form m + n\/-2?
8.11. Prove that the number of primes less than or equal to JV is at least log 2 (iV/3),
by proceeding as follows. Let p\,... ,pn be the prime numbers among 1,.. .,N, and
let È(Í) be the number of square-free integers among 1,..., JV, that is, the integers
not divisible by any square number. We then have the following relation, since it
is known that ÓÃ=é(^1 /^^2 ) = 71-2/6-
ç
È(Í) > Í-Ó
fc=l
(Here the square brackets denote the greatest-integer function.) Now a square-free
integer k between 1 and Í is of the form k = pj^1 · · · p£", where each e, is either 0 or
- Hence È(Í) < 2 n, and so ç > log 2 (iV/3). This interesting bit of mathematical
trivia is due to the Russian-American mathematician Joseph Perott (1854-1924).
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8.12. Assuming that lim exists, use Chebyshev's estimates to show
ç—>oo ç
that this limit must be 1 and hence that Legendre's estimate cannot be valid beyond
the first term.