Find the maximum value of a positive integer m, such that m
3+8 is a perfect square.
for m=1:99999
sq=m^3+8;
sr=round(sqrt(sq));
if sr^2==sq
disp([m sq sr])
end
end
for m=sym(100000):2000000
sq=m^3+8;
sr=round(sqrt(sq));
if sr^2==sq
disp([m sq sr])
end
end
finds only
m m^3+8 sq root
1 9 3
2 16 4
46 97344 312
so it appears that 46 is the maximum such m.
Edited on December 18, 2023, 12:58 pm
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Posted by Charlie
on 2023-12-18 12:38:20 |
computer solution