algebraic equation

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∵ a < b < c < d are positive integer. ∴
4/d < 1/a + 1/b + 1/c + 1/d = 1 < 4/a
⇒
a < 4 < d
When a = 1 ,
1/b + 1/c + 1/d = 0 is impossible since b , c , d are positive integer.
When a = 3 ,
1/a + 1/b + 1/c + 1/d ≤ 1/3 + 1/4 + 1/5 + 1/6
⇒
1 ≤ 19/20 (contradiction)
Hence a = 2 ,
i.e.
1/b + 1/c + 1/d = 1/2 ,
then 3/d < 1/2 < 3/b
⇒
b < 6 < dWhen (a , b) is (2 , 3) ,
1/2 + 1/3 + 1/c + 1/d = 1
1/c + 1/d = 1/6
6c + 6d = cd
6d = c(d - 6)
6(d - 6) = c(d - 6) - 36
(c - 6)(d - 6) = 36For 6 < d ,
(c - 6) (d - 6) = 1 * 36 or 2 * 18 or 3 * 12 or 4 * 9
⇒
(c , d) = (7 , 42) or (8 , 24) or (9 , 18) or (10 , 15)
When (a , b) is (2 , 4) ,
1/2 + 1/4 + 1/c + 1/d = 1
1/c + 1/d = 1/4
4c + 4d = cd
4c = d(c - 4)
4(c - 4) = d(c - 4) - 16
(c - 4)(d - 4) = 16For 6 < d ,
(c - 4)(d - 4) = 1 * 16 or 2 * 8
⇒
(c , d) = (5 , 20) or (6 , 12)
When (a , b) is (2 , 5) ,
1/2 + 1/5 + 1/c + 1/d = 1
1/c + 1/d = 3/10
10c + 10d = 3cd
10c = d(3c - 10)
30c = 3d(3c - 10)
10(3c - 10) = 3d(3c - 10) - 100
(3c - 10)(3d - 10) = 100For c > b = 5 ,
3c - 10 = 3*6 - 10 = 8 only since 3*7 - 10 = 11 > √100.
No solution in this case since 8 is not a factor of 100.
By the result above , there are 6 sets (a,b,c,d) for 1/a + 1/b + 1/c + 1/d = 1 :1/2 + 1/3 + 1/7 + 1/42 = 1
1/2 + 1/3 + 1/8 + 1/24 = 1
1/2 + 1/3 + 1/9 + 1/18 = 1
1/2 + 1/3 + 1/10 + 1/15 = 1
1/2 + 1/4 + 1/5 + 1/20 = 1
1/2 + 1/4 + 1/6 + 1/12 = 1