Euler Conjecture and CDC 6600

https://fortran-lang.discourse.group/t/euler-conjecture-and-cdc-6600/10501

Comments

nroetsDec 4, 2025, 5:10 PM
Here's a fairly fast algorithm: Look for a number that appears in both { j⁵+k⁵+l⁵ } and { i⁵-n⁵ } where 0 < j <= k <= l and 0 < n < i

If we only consider l and i under 250, then the sets would contain less than 3 million integers each.

Strength reduction can be used to replace all the multiplications with additions: https://en.wikipedia.org/wiki/Strength_reduction

asplakeDec 4, 2025, 3:10 PM
Re precomputing fifth powers, seems Fortran not only has array comprehensions, but compile-time array comprehensions. It was never exactly my cup of tea, but nearly 40 years out of university it seems Fortran has kinda kept up!
zaikunzhangDec 4, 2025, 3:50 AM
See also

https://community.intel.com/t5/image/serverpage/image-id/116...

jmclnxDec 4, 2025, 1:16 PM
FWIW on my W541, Slackware I get

i^5 = 61917364224 j^5 + k^5 + l^5 + n^5 = 61917364224 27 84 110 133 144

Slackware W541:

real 1m6.703s

user 1m6.658s

sys 0m0.002s

Output is the same, which is aloways good. But times are slower because I am on a ~10 year old system :)

So I gave it a whirl on sdf.org's Debian system Debian 6.1.140-1 (2025-05-22) x86_64 GNU/Linux kenel 6.1.0-37-amd64

real 1m9.615s

user 1m9.595s

sys 0m0.016s

That system is a true multi user system and had 54 users logged in when I ran it. So I think it is better than I expected.

FWIW, I believe my first paid job was on a 6600 while in college, but it could have been a 7600. There were upgrading the system from the 6600 when I was there.

nurettinDec 4, 2025, 5:31 AM
I like the bald guy in the comments idea golfing and posting his own numbers.
SomeoneDec 4, 2025, 6:00 AM
For a limit of 10,000⁵. Reading https://www.ams.org/journals/mcom/1967-21-097/S0025-5718-196... the original search seems to have gone up to 250⁵. That’s a search space that’s 40⁵ ≈ 10⁸ larger.

They also forget to break out of the loops when the sum of, say, the first three fifth powers already is larger than 10,000⁵.

nroetsDec 4, 2025, 6:25 AM
It's likely that the original search also used strength reduction to save a lot of cycles (effectively replacing all multiplications with additions):

https://en.wikipedia.org/wiki/Strength_reduction