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

By Euler L.

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From the PREFACE. THE ebook of this tract has been behind schedule by means of quite a few explanations, and i'm now forced to factor it with out Dr Riesz's assist in the ultimate correction of the proofs. This has at any cost, one virtue, that it provides me the possibility of asserting how wide awake i'm that no matter what worth it possesses is due almost always to his contributions to it, and particularly to the actual fact, that it includes the 1st systematic, account of his appealing idea of the summation of sequence through 'typical means'.

Extra resources for 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

Example text

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.