We need only test postive x, but for every such (x,y) solution, (-x,y) will also be a solution.
We need only test non-negative y since negative y would yield non integer results.
A computer search shows that the only solutions are (0,0), (2,1), and (-2,1).
Analytical approach:
x^4 + x^2 - 21^y-1 = 0
x^2 = (1/2){-1 + sqrt(4*(21^y)-3)}
(would be +/- but the '-' option gives imaginary x)
So 4*(21^y)-3 must be a perfect square and must be odd.
if y=0, 4*(21^y)-3 = 1 and x=0
if y=1, 4*(21^y)-3 = 81 and x= +/- 2
Note that perfect squares, modulo 9, can only be [0,1,4,7].
Consider that for k in [1,2,3,...] 21*k, modulo 9, is [3,0,0,0,...]
And 4*21^y-3 (modulo 9) is 1 for y=0 and is 0 for y=1.
But for all integers y larger than 1, it is 6.
Perfect squares, modulo 9, can never be 6.
So no other y will give an integer result.
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Posted by Larry
on 2021-01-03 10:07:42 |
Computer and Analytic Solution
