If
(√6 + √2)9 A + B√C
---------- = -------
(√8)9 D
Then, find the integers values of A, B, C, and D with C square-free.
(In reply to
computer solution by Charlie)
I realize I misunderstood the definition of square-free. The program should have been
clc
lhs=(sqrt(6)+sqrt(2))^9/(8192*sqrt(2))
for c=1:50
f=factor(c)';
f=groupcounts(f);
f=f<2;
f=all(f);
if all(f)
for a=1:500
for b=1:500
for d=1:5000
if abs((a+b*sqrt(c))/d-lhs)<.0000001
fprintf('%4d%4d%4d%4d %18.14f %18.14f\n',[a b c d (a+b*sqrt(c))/d (a+b*sqrt(c))/d-lhs])
end
end
end
end
end
end
lhs =
16.5626179236887
265 153 3 32 16.56261792368869 0.00000000000001
489 172 7 57 16.56261799128263 0.00000006759394
219 301 42 131 16.56261794061653 0.00000001692785
262 335 46 153 16.56261793690827 0.00000001321959
showing (A,B,C,D) = (265, 153, 3, 32) as a unique solution, as 27 is not square-free -- it's divisible by 9.
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Posted by Charlie
on 2025-03-17 10:51:22 |
re: computer solution
