Euler Conjecture and Cdc 6600

I like the bald guy in the comments idea golfing and posting his own numbers.

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.

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 n 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

See also

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

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!