All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Reciprocal Equation #2 (Posted on 2003-07-20)
Find all sets of positive integers A, B, and C which satisfy

1/A = 1/B + 1/C.

 See The Solution Submitted by Brian Smith Rating: 3.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 hmmm... | Comment 1 of 15
There doesn't seem to be a limit as to the number of sets that satisfy the equation.

The equation becomes
bc=ac+ab, or
bc=a(b+c), or
a = bc/(b+c)

The following program seems to produce solutions going on and on:
```
DECLARE FUNCTION gcd# (x#, y#)

DEFDBL A-Z

sum = 1

DO

sum = sum + 1

FOR b = 1 TO sum / 2

c = sum - b

a = b * c / sum

IF a = INT(a) THEN

IF gcd(gcd(a, b), c) = 1 THEN

PRINT a, b, c, sum

END IF

END IF

NEXT

LOOP

FUNCTION gcd (x, y)

dnd = x: dvr = y

IF dnd < dvr THEN SWAP dnd, dvr

DO

q = INT(dnd / dvr)

r = dnd - q * dvr

dnd = dvr: dvr = r

LOOP UNTIL r = 0

gcd = dnd

END FUNCTION

such as:

1             2             2

2             3             6

3             4             12

4             5             20

6             10            15

5             6             30

6             7             42

10            14            35

12            21            28

7             8             56

15            24            40

8             9             72

14            18            63

20            36            45

9             10            90

21            30            70

10            11            110

18            22            99

24            33            88

28            44            77

30            55            66

11            12            132

35            60            84

12            13            156

...

102           138           391

112           161           368

120           184           345

126           207           322

130           230           299

132           253           276

23            24            552

95            120           456

119           168           408

143           264           312

...

330           555           814

336           592           777

340           629           740

342           666           703

37            38            1406

105           114           1330

165           190           1254

217           266           1178

261           342           1102

297           418           1026

325           494           950

345           570           874

357           646           798

...

69            70            4830

201           210           4690

549           630           4270

649           770           4130

741           910           3990

901           1190          3710

969           1330          3570

1081          1610          3290

1161          1890          3010

1189          2030          2870

1209          2170          2730

1221          2310          2590

70            71            4970

138           142           4899

204           213           4828

268           284           4757

330           355           4686

390           426           4615

448           497           4544

504           568           4473

558           639           4402

610           710           4331

660           781           4260

708           852           4189

754           923           4118

798           994           4047

840           1065          3976

880           1136          3905

918           1207          3834

954           1278          3763

and so forth.

Maybe there's a pattern here, and solutions can be categorized, so as to be able to enumerate the classes.  The program already weeds out those where a, b and c share a common factor, such as 2,4,4, which is merely a reiteration of 1,2,2.

Posted by Charlie
on 2003-07-20 04:49:28

```

 Search: Search body:
Forums (0)