Euler's sum of powers conjecture: Difference between revisions

mantissa. 249^4 is not too big to fit in those 4 bytes, but even 215^4 fills up all available bits. Adding up logarithmic "percentages" by subtracting that of the target sum's square root from that of each summand and seeing if the inverses add up to the square root, will still lose precision and find false solutions, as a difference of 2 in one of the summands results in a difference in just the 11th place after the decimal point. Even QL SuperBASIC can only makemakes such FP comparisons down to the 7th. But the attempted fix should only be very slow.}}

25 DIM p(249): DIM q(249)

415 FOR i=1 TO 249: q(i)=5*LN i: p(i)=q(i)/2: NEXT i

20 FOR w=7 TO 245 STEP 7: FOR x=5 TO 245 STEP 5: FOR y=3 TO 246 STEP 3: FOR z=2 TO 248 STEP 2

30 LET sum=EXP(q(w)-p(m))+ EXP(q(x)-p(m)) + EXP(q(y)-p(m)) + EXP(q(z)-p(m))

35 LET lnsum=LN sum

60 NEXT z: NEXT y: NEXT x: NEXT w