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.
clc
lhs=(sqrt(6)+sqrt(2))^9/(8192*sqrt(2))
for c=1:50
f=factor(c)';
f=groupcounts(f);
f=mod(f,2);
f=all(f); %square free c
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
finds
LHS = 16.5626179236887
A B C D RHS Diff
265 153 3 32 16.56261792368869 0.00000000000001
489 172 7 57 16.56261799128263 0.00000006759394
13 499 8 86 16.56261785172499 -0.00000007196369
50 341 27 110 16.56261796493601 0.00000004124733
265 51 27 32 16.56261792368869 0.00000000000001
315 392 27 142 16.56261795564084 0.00000003195216
26 499 32 172 16.56261785172499 -0.00000007196369
197 380 40 157 16.56261797278961 0.00000004910093
219 301 42 131 16.56261794061653 0.00000001692785
262 335 46 153 16.56261793690827 0.00000001321959
The first and fifth lines seem to be solutions. Taking the first and verifying to more decimal places:
>> (sqrt(vpa(6))+sqrt(vpa(2)))^9/(sqrt(vpa(8)))^9
ans =
16.562617923688694559678102820325
>> (vpa(265)+vpa(153)*sqrt(vpa(3)))/vpa(32)
ans =
16.562617923688694559678102820325
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Posted by Charlie
on 2025-03-17 09:42:13 |
computer solution
