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380 Chapter 7 ELLIPTIC CURVE ARITHMETIC<br />

7.22. For the example prime p = 2 31 +1 /3 and its curve orders displayed<br />

after Algorithm 7.5.10, which is the best order to use to effect an ECPP proof<br />

that p is prime?<br />

7.23. Use some variant of ECPP to prove primality of every one of the ten<br />

consecutive primes claimed in Exercise 1.87.<br />

7.24. Here we apply ECPP ideas to primality testing of Fermat numbers<br />

Fm =2 2m<br />

+ 1. By considering representations<br />

4Fm = u 2 +4v 2 ,<br />

prove that if Fm is prime, then there are four curves (mod Fm)<br />

y 2 = x 3 − 3 k x; k =0, 1, 2, 3,<br />

having, in some ordering, the curve orders<br />

2 2m<br />

+2 m/2+1 +1,<br />

2 2m<br />

− 2 m/2+1 +1,<br />

2 2m<br />

− 1,<br />

2 2m<br />

+3.<br />

Prove by computer that F7 (or some even larger Fermat number) is composite,<br />

by exhibiting on one of the four curves a point P that is not annihilated by any<br />

of the four orders. One should perhaps use the Montgomery representation<br />

in Algorithm 7.2.7, so that initial points need have only their x-coordinates<br />

checked for validity (see explanation following Algorithm 7.2.1). Otherwise,<br />

the whole exercise is doomed because one usually cannot even perform squarerooting<br />

for composite Fm, to obtain y coordinates.<br />

Of course, the celebrated Pepin primality test (Theorem 4.1.2) is much<br />

more efficient in the matter of weeding out composites, but the notion of CM<br />

curves is instructive here. In fact, when the above procedure is invoked for<br />

F4 = 65537, one finds that indeed, every one of the four curves has an initial<br />

point that is annihilated by one of the four orders. Thus we might regard<br />

65537 as a “probable” prime in the present sense. Just a little more work,<br />

along the lines of the ECPP Algorithm 7.5.9, will complete a primality proof<br />

for this largest known Fermat prime.<br />

7.8 Research problems<br />

7.25. With a view to the complexity tradeoffs between Algorithms 7.2.2,<br />

7.2.3, 7.2.7, analyze the complexity of field inversion. One looks longingly at<br />

expressions x3 = m 2 − x1 − x2, y3 = m(x1 − x3) − y1, in the realization that<br />

if only inversion were “free,” the affine approach would surely be superior.<br />

However, known inversion methods are quite expensive. One finds in practice<br />

that inversion times tend to be one or two orders of magnitude greater than

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