A solution to a problem of Fermat, on two numbers of which - download pdf or read online

By Euler L.

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D d Proof. By B´ezout, there are integers r, s with rm + sn = d. Dividing both members by d gives n m r + s = 1, d d and the result is immediate. 17 Theorem (Order of a Power). Still modulo n > 0. Suppose ordn (a) = d. Then ordn (ak ) = Proof. considering classes d . (d, k) By our Main Theorem (ak )e ≡ 1 (mod n) ⇐⇒ d|ke ⇐⇒ d k |e · . (d, k) (d, k) As d/(d, k) and k/(d, k) are relatively prime by the Lemma, our First Divisibility Theorem then shows that (ak )e ≡ 1 (mod n) ⇐⇒ d |e. (d, k) So the least positive e with the property stated in the left member is e = d/(d, k).

28 CHAPTER A. 18 Example. There are two extreme cases, (d, k) = 1 and k|d. For instance, modulo 13 we have verified that ordn (2) = 12. The Theorem then shows that the classes [2]5 = [6], [2]7 = [11], [2]11 = [7] have that same order, as the exponents are relatively prime to 12. The class of [2]6 = [64] = [−1] has order 2 = 12/6 as predicted by the Theorem. We have exemplified the fact that mutually inverse classes have the same order – we leave the simple general proof as an exercise. A natural question regards the order of a product.

For composite N a similar result holds. , if neither 2 nor 5 divides N . The period d equals the order of 10 modulo N and is a factor in φ(N ). If (10, N ) > 1 the decimal expansion cannot be purely periodic. By the reasoning above, the period d would entail 10d ≡ 1 (mod N ). But if ad ≡ 1 (mod N ), then clearly a and ad−1 are inverse to one another modulo N . And invertibility means (a, N ) = 1. Let us look at an example, N = 840 = 2 · 3 · 4 · 5 · 7. 001190476190476109 . . It has a preperiod of three places, 001, followed by a periodic part, of period 6.

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A solution to a problem of Fermat, on two numbers of which the sum is a square and the sum of their squares is a biquadrate, inspired by the Illustrious La Grange by Euler L.


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