Question

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7. Consider the generalized Pell equation x2- 6y2 =3. *) Verify that 3+ v6 E Z[V6] gives a solution of (*). Produce three other non-trivial positive solutions of (. (Hint: Use solutions of x2 - 6/2 = 1 to help out.)

Answer 1

Clearly, 3+√6 is a solution of x² - 6y² = 3 as, 3² - 6.1² = 3

Note the following theorem :

Fix u = a + bVd where a2 - db2 Theorem = 1 with a and b in Z+. For each {0}, every solution of x2 - dy2 Jaæ VInu and |y| < V[2/u/Vd. = n is a Pell multiple of a solution (x,y) where n E Z-

We will use the results of this theorem in the following main proof.

We will describe all the solutions of x2 - 6y2 solution is (3, 1) and we'll show every solution is a Pell multiple of (3,1) positive solution of a2 - 6b2 =1 we take (a, b) = (5,2), the bound y VIn u/Vd = /3u/V = 3 in integers. An obvious or (-3, -1 so set u 5+2/6. From 2 1.25 forces y to be 1,0, or -1. For a Theorem Solutions to a2 - 6y2 tells us that every solution of 2- 6y2 = 3 in integers has the form = 3 with such y-values are (±3,1) and (±3, -1). Therefore Theorem (+3- V6)(5+2V6) + yV6= (+3 + v©)(5+2 /6) k or where k E Z. Up to a Pell multiple there are four possible solutions: V6, -3 V6 -36, 36, 3- (3 - V6)u. Likewise 3 in integers is a Pell The solutions 3 V6 and 3 - V6 V6 (-3V6)u. Therefore every solution of x2 - 6y2 multiple of +(3 6). Thus every solution has the form ±(3+ v6)(52/6)k with kE Z and we can't simplify this further since t(3 /6) Taking k 0,1,2, the values of (3+v6)(5+2/6) so the first three solutions of x2 - 6y2 (267, 109) are Pell multiples: 3 + -3- are not Pell multiples of each other. are 3+V6, 27+11/6, and 267+ 109 /6, (3,1), (27, 11), and = 3 in positive integers are