Determine the
minimum value of a prime number p such that there are positive integers
x and
y satisfying this equation:
x2 + p3 = y4
*** As an extra challenge only, solve this puzzle without using a computer program/spreadsheet aided method.
pr=2;
while 1==1
cube=pr^3;xs=2;
for x=1:100000
prev=xs;
xs=x^2;
p4=xs+cube;
y=round(sqrt(sqrt(p4)));
if y^4==p4
disp([x xs pr pr^3 y p4]);
end
if xs-prev>cube
break
end
end
pr=nextprime(pr+1);
end
finds
6083 37002889 23 12167 78 37015056
which is
6083^2 + 23^3 = 78^4
or, with its powers evaluated,
37002889 + 12167 = 37015056
The answer is that 23 is the smallest prime that fits the bill.
Judging by the long time with no further solutions, it might be that this is the only prime that will work, not just the smallest. (Tested up to p=6551)
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Posted by Charlie
on 2022-08-14 10:50:14 |
computer solution
