The great Euler conjected,
inter alia, that at least n powers of positive integers are needed to get a sum which is a n-th power as well.
It is up to you to find a set of integers {a,b,c,d,e}
such that
a^5+b^5+c^5+d^5=e^5.
REM: If unable to provide a proof, present the current state of known solutions for 4th and 5th powers.
AFAIK no solution for the 6th power exists so far.
clearvars,clc
upto=500;
for i=1:upto
p5(i)=i^5;
end
mx=p5(upto);
for a=1:upto-4
t1=p5(a);
for b=a+1:upto-3
t2=t1+p5(b);
if t2>mx
break
end
for c=b+1:upto-2
t3=t2+p5(c);
if t3>mx
break
end
for d=c+1:upto-1
t4=t3+p5(d);
if t4>mx
break
end
tst=round(t4^(1/5));
if tst^5==t4
disp([a b c d tst]);
end
end
end
end
end
finds
a b c d e
27 84 110 133 144
54 168 220 266 288
81 252 330 399 432
where the second and third lines are merely twice and thrice the first. Any further multiples have e exceeding the maximum sought.
Edited on July 6, 2022, 12:05 pm
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Posted by Charlie
on 2022-07-06 12:04:20 |
computer solution
