- #31
mfb
Mentor
- 37,391
- 14,221
It generates a large number of solutions, but not all:
n= 35, a= 5 , b= 21 c= 28
As three numbers need a common divisor, we have m=7 as only option (m=1 doesn't fit with the generating equations). This leads to k=5 and c=1:
(7*5 ; 3*7 , 4*7 , 1*5)
We also need 7=1+2*3 (correct) and 1=(5-4)(5-3) (wrong).
k=a+b, k=a+a and k=b+b lead to the trivial solutions.
n= 35, a= 5 , b= 21 c= 28
As three numbers need a common divisor, we have m=7 as only option (m=1 doesn't fit with the generating equations). This leads to k=5 and c=1:
(7*5 ; 3*7 , 4*7 , 1*5)
We also need 7=1+2*3 (correct) and 1=(5-4)(5-3) (wrong).
k=a+b, k=a+a and k=b+b lead to the trivial solutions.
Code:
15 5 5 12
15 10 10 3
20 5 5 18
20 15 15 2
35 7 14 30
42 7 28 30
55 11 33 40
63 9 27 56
63 21 28 45
65 26 26 45
65 39 39 20
66 11 22 60
72 16 24 63
77 11 44 63
78 13 13 75
85 17 17 80
90 27 36 70
99 11 44 90
112 16 32 105
136 51 51 100
152 19 57 140
175 75 75 112
203 58 58 175
259 37 37 252
290 29 116 270
310 62 93 280