By Zariski O.

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Continuing in our optimism: Now we hope that we can adjust the solution 10 mod 72 by adding some multiple of 72 to it in order to get a solution mod 73. That is, we hope to nd y so that 10 + 72 y2 2 mod 73 Multiplying out and simplifying, this is Dividing through by 72 gives 294y ,98 mod 73 6y ,2 mod 7 Again, the inverse of 6 mod 7 is just 6 again, so this is y 6,2 2 mod 7 Therefore, satis es 10 + 72  2 = 108 1082 2 mod 73 This was considerably faster than brute-force hunting for a square root of 2 mod 73 directly.

The next integer after 2 on the list which hasn't been marked is 3. Starting with 3 + 3, mark every 3rd integer counting those already marked. This marks all multiples of 3 bigger than 3 itself. The next integer after 3 on the list which hasn't been marked is 5. Starting with 5 + 5, mark every 5th integer counting those already marked.  ... Take the next integer n on the list which has not yet been crossed-o . This n is prime. Starting with n + n, cross o every nth integer counting those already marked.

And, funnily enough, the procedure to do so is exactly parallel to Newton's method for numerical approximation to roots, from calculus. In particular, we will use a purely algebraic form of Taylor expansions to prove the result. First we'll do a numerical example to illustrate the idea of the process to which Hensel's Lemma refers. Suppose we want to nd x so that x2 2 mod 73. Noting that a solution mod 73 certainly must give a solution mod 7, we'll start by nding a solution mod 7. This is much easier, since there are only 7 things in Z=7, and by a very quick trial and error hunt we see that 32 = 9 = 2 mod 7.

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