Show that if (2 + √3)k = 1 + m + n√3, for positive integers m, n, k with k odd, then m is a perfect square.
Solution
We have (2 + √3)4 = 97 + 56√3 = 14 (7 + 4√3) - 1 = 14 (2 + √3)2 - 1. Hence (2 + √3)k+2 = 14 (2 + √3)k - (2 + √3)k-2. Thus if (2 + √3)k = ak + bk√3, then ak+2 = 14 ak - ak-2.
Now suppose the sequence ck satisfies c1 = 1, c2 = 5, ck+1 = 4 ck - ck-1. We claim that ck2 - ck-1ck+1 = 6. Induction on k. We have c3 = 19, so c22 - c1c3 = 25 - 19 = 6. Thus the result is true for k = 2. Suppose it is true for k. Then ck+1 = 4 ck - ck-1, so ck+12 = 4 ckck+1 - ck-1ck+1 = 4 ckck+1 - ck2 + 6 = ck(4 ck+1 - ck) + 6 = ckck+2 + 6, so the result is true for k+1.
Now put dk = ck2 + 1. We show that dk+2 = 14 dk+1 - dk. Induction on k. We have d1 = 2, d2 = 26, d3 = 362 = 14 d2 - d1, so the result is true for k = 1. Suppose it is true for k. We have ck+3 - 4 ck+2 + ck+1 = 0. Hence 12 + 2 ck+3ck+1 - 8 ck+2ck+1 + 2 ck+12 = 12. Hence 2 ck+22 - 8 ck+2ck+1 + 2 ck+12 = 12. Hence 16 ck+22 - 8 ck+2ck+1 + ck+12 + 1 = 14 ck+22 + 14 - ck+12 - 1, or (4 ck+2 - ck+1)2 + 1 = 14 (ck+22 + 1) - (ck+12 + 1), or ck+32 + 1 = 14 (ck+22 + 1) - (ck+12 + 1), or dk+3 = 14 dk+2 - dk+1. So the result is true for all k.
But a1 = 2, a3 = 26 and a2k+3 = 14 a2k+1 - a2k-1, and d1 = 2, d2 = 26 and dk+1 = 14 dk - dk-1. Hence a2k-1 = dk = ck2 + 1.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002