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Concrete mathematics : a foundation for computer science

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I have discovered a<br />

wonderful proof of<br />

Fermat’s Last Theorem,<br />

but there’s no<br />

room <strong>for</strong> it here.<br />

There<strong>for</strong>e, if Fermat’s<br />

Last Theorem<br />

is false, the universe<br />

will not be big<br />

enough to write<br />

down any numbers<br />

that disprove it.<br />

A ANSWERS TO EXERCISES 509<br />

Incidentally, this exercise implies that (m + m”)/(n + n”) = m//n’,<br />

although the <strong>for</strong>mer fraction is not always reduced.<br />

4.62 2 ‘$2 2+2 3-2 6-2 7+2~'2+2~~'3-2~20-2~21+2~30+2~31-<br />

2 ~42 - 2 43 + . . . can be written<br />

; + 3 t(2-4k1-6k-3 _ 2-4k2-10k -7)<br />

k>O<br />

Incidentally, this sum can be expressed in closed <strong>for</strong>m using the “theta function”<br />

O(z, h) = tk e~xhkz+2irk; we have<br />

e t-3 i + ~;6(~ln2,3iln2) - &O(%ln2,5iln2)<br />

4.63 Any n > 2 either has a prime divisor d or is divisible by d = 4. In either<br />

case, a solution with exponent n implies a solution (an/*)*+(bn/*)* = (c”/*)*<br />

with exponent d. Since d = 4 has no solutions, d must be prime.<br />

The hint follows from the binomial theorem, since aP+(x-a)P-pap ’<br />

is a multiple of x when p is odd. Assume that a -L x. If x is not divisible<br />

by p, x is relatively prime to cP/x; hence x = mp <strong>for</strong> some m. If x is divisible<br />

by p, then cp/x is divisible by p but not by p2, and cp has no other factors<br />

in common with x.<br />

(The values of a, b, c must, in fact, be even higher than this result<br />

indicates! Inkeri [160] has proved that<br />

A sketch of his proof appears in [249, pages 228-2291, a book that contains<br />

an extensive survey of progress on Fermat’s Last Theorem.)<br />

4.64 Equal fractions in YN appear in “organ-pipe order”:<br />

2m 4m rm 3m m<br />

- - - -<br />

2n’ 4n’ . . . . ml . . . . 3n’ n.<br />

Suppose that IPN is correct; we want to prove that &+I is correct. This<br />

means that if kN is odd, we want to show that<br />

k-l<br />

N+l<br />

= yN,kN;<br />

if kN is even, we want to show that<br />

k-l<br />

yN,kN 1 yN,kN ~ N+l<br />

yN,kN YN,kN+l *

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