0,1

These appear to be the only solutions to a! + b! + c! + d! = n^2:

a b c d n

0 0 0 0 4

0 0 0 1 4

0 0 0 3 9

0 0 1 1 4

0 0 1 3 9

0 1 1 1 4

0 1 1 3 9

0 2 3 6 729

0 4 4 5 169

0 4 8 9 403225

0 5 5 5 361

0 5 5 6 961

0 5 7 7 10201

1 1 1 1 4

1 1 1 3 9

1 2 3 6 729

1 4 4 5 169

1 4 8 9 403225

1 5 5 5 361

1 5 5 6 961

1 5 7 7 10201

2 2 3 3 16

2 2 6 6 1444

4 5 9 9 725904

1!+2!+3!+6! = 729 = 27^2. This shows that 4 factorials can add to a cube.

e = 75; a = Union[ Flatten[ Table[a! + b! + c! + d!, {a, 1, e}, {b, a, e}, {c, b, e}, {d, c, e}]]]; l = Length[a]; Do[ If[ IntegerQ[ Sqrt[ a[[i]] ]], Print[ a[[i]] ]], {i, 1, l}]

(PARI) sum4factsq(n) = { for(a1=0, n, for(a2=a1, n, for(a3=a2, n, for(a4=a3, n, z = a1!+a2!+a3!+a4!; if(issquare(z), print(a1" "a2" "a3" "a4" "z)) ) ) ) ) }

Cf. A082875.

Adjacent sequences: A082917 A082918 A082919 this_sequence A082921 A082922 A082923

Sequence in context: A038784 A038239 A089149 this_sequence A110979 A073173 A092464

easy,nonn

Cino Hilliard (hillcino368(AT)gmail.com), May 25 2003

Edited, corrected and extended by Robert G. Wilson v (rgwv(AT)rgwv.com), May 26 2003