By Peter J. Cameron

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**Extra resources for A Course on Number Theory [Lecture notes]**

**Example text**

This completes the proof. 1 Find the first six terms in the continued fraction for the number e, the root of natural logarithms. ) Note: You might spot a pattern here. The pattern really does continue! 2 Find the continued fractions for the following numbers: √ (a) 3 + 2 2; √ (b) 11 − 10; √ (c) (1 + 5)/4. 3 Why does the number (−1 + −3)/2 not have a continued fraction expansion? 4 Let Fn be the nth Fibonacci number, defined by the rules F1 = 1, F2 = 2, Fn = Fn−1 + Fn−2 for n ≥ 3. (a) Show that [1, 1, .

So the result is true for k = 0. For k > 0, we have √ n = [a0 ; a1 , . . , ak , yk+1 ] yk+1 pk + pk−1 = yk+1 qk + pk √ Pk+1 pk + Qk+1 pk−1 + pk n √ , = Pk+1 qk + Qk+1 qk−1 + qk n √ using yk+1 = (Pk+1 + n)/Qk+1 . Hence √ n(Pk+1 qk + Qk+1 qk−1 − pk ) = Pk+1 pk + Qk+1 pk−1 − qk n. √ Since n is irrational, and everything else in the equation is an integer, both sides must be zero. 7)(a) from Chapter 3. 7 Suppose that pk /qk is the kth convergent to n. Then p2k −nq2k = ±1 √ if and only if k is one less than a multiple of the period of the continued fraction for n.

Al−1 , y + a0 ] (y + a0 )pl−1 + pl−2 = , (y + a0 )ql−1 + ql−2 where pk /qk are convergents to y + a0 . Now 2a0 ql−1 + ql−2 = 2a0 [a1 , . . , al−1 ] + [a1 , . . , al−2 ] = 2a0 [a1 , . . , al−1 ] + [a2 , . . , al−1 ] = pl−1 , 56 CHAPTER 6. LAGRANGE AND PELL using the fact that [a1 , . . , al−2 ] = [al−1 , . . , a2 ] = [a2 , . . , al−1 , ] the first equality because a1 = al−1 , . . 5)(a)) in Notes 2). Hence y + a0 = (y + a0 )pl−1 + pl−2 . (y − a0 )ql−1 + pl−1 Multiplying up and cancelling, we obtain (y2 − a20 )ql−1 = pl−2 , øy = √ r where r = a20 + pl−2 /ql−1 , as claimed.

### A Course on Number Theory [Lecture notes] by Peter J. Cameron

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