Exercises 41
5n7d13

(AB)dJ
2d3d5n−d11
(EF
)dK
2d5n−d7d13

(AB)dJ
22d3d5n−2d11
(EF
)dK
22d5n−2d7d13

(AB)dJ
.
.
.
(EF
)dK
2qd5r7d13

(AB)rA
2n3r7d−r−117
(if r 0) C I (if r = 0)
2n3r−17d−r−119 2n7d−1

(DG)nH

LnM d−1N
3r−15n7d−r11 3n5n+111
(EF
)r−1K
(EF
)nK
5n7d−113 5n+17n13
Figure 4. The action of Conway’s prime-producing machine when
started with
5n7d13,
where 0 d n. The variables q and d are
defined by the division algorithm: n = dq + r where 0 r d.
every pair a, b S and every odd prime p. We make this assumption
from now on.
(c) Now argue that v2(a) = v2(b) for every pair of elements a, b S.
Thus, dividing through by a suitable power of 2, we may (and do)
assume that all the elements of S are odd.
(d) Finally, show that for each pair of elements a, b S, we have
a + b =
2v2(a+b)
p2
pmin{vp(a),vp(b)}.
Show that this equation leads to a contradiction if a and b are
chosen to be congruent modulo 4.
24. Figure 4, based on Conway’s article [Con87], describes the action of
Conway’s prime-producing machine. Decipher this figure and explain
how it proves Theorem 1.8. For a more detailed explanation of the
workings of Conway’s prime-producing machine, see Guy’s expository
article [Guy83].
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