[Guest]

To Thalia and other consultants in the forum:

Is there a formula to express a given number as the sum of the squares of three positive integers? I am looking for three ways to express 1989 as the sum of the squares of three positive integers. There are simply too many squared numbers to try out and I am going now where with any of them. Looking for hints and formula if there are any out there. Thanks.

[Thalia]

Does the problem just say 3 positive integers or does it specify consecutive or multiples?

[Guest]

Does the problem just say 3 positive integers or does it specify consecutive or multiples?

Thalia, Hiiiii. It only says "3 positive integers". I'm hoping that you will come up with some hints for me.

[Thalia]

Well, the formula would be a^2+b^2+c^2=1989. I have no clue how you would solve that though.

[Guest]

Look at it by solving for each letter.

a^2 = 1989 - b^2 - c^2 = (1989 -b^2 - c^2) + b^2 + c^2 = 1989 (Just sovele for the rest of the letters and plug those in till you just have c^2 to deal with

Hope I'm helpful. Algebra is one of my favorite things to do

Edited February 9, 2011 by Darth Legion

[Thalia]

Maybe I'm missing something but doesn't that give you 1989=1989?

[Guest]

(1989 -[-(1989 -[-a^2 - c^2 +1989] -c^2) -c^2 +1989] -c^2) +c^2 + (-[1989 - b^2 - c^2] -c^2 +1989) = 1989

sorry didn't think it through all the way....

[Guest]

Maybe I'm missing something but doesn't that give you 1989=1989?

No, no, you are right. (It was early, I was cold, I was tired ect ect) I messed up lol

[Guest]

I got an exact answer using a square root calculator the goes up to something like 20 decimals, but one of the numbers is 18 decimals. I'm going to find out if there is even a possible answer other than the formula.

BTW I got it by reverse engineering aka. 1600 + 300 + 89 (the squared amounts)

Edited February 9, 2011 by Darth Legion

[curr3nt]

I got an exact answer using a square root calculator the goes up to something like 20 decimals, but one of the numbers is 18 decimals. I'm going to find out if there is even a possible answer other than the formula.

BTW I got it by reverse engineering aka. 1600 + 300 + 89 (the squared amounts)

89 and 300 are the square of what integer?

Also Darth, I think you are not supposed to give actual answers in this area.

Not exactly an eligant approach but you can determine that none of the numbers are 45 or greater since the square root of 1989 is 44.598. So you can make 44 equations of A^2 + B^2 = 1989 - C^2 using the values 1-44 for C. Might be easier to spot then.

edit - you do much geometry? hint: think of a right triangle with integer sides. I found two with this.

Edited February 9, 2011 by curr3nt

[Guest]

89 and 300 are the square of what integer?

Also Darth, I think you are not supposed to give actual answers in this area.

Not exactly an eligant approach but you can determine that none of the numbers are 45 or greater since the square root of 1989 is 44.598. So you can make 44 equations of A^2 + B^2 = 1989 - C^2 using the values 1-44 for C. Might be easier to spot then.

edit - you do much geometry? hint: think of a right triangle with integer sides

LOL actually it's a really long float value

[curr3nt]

Found a third set of integers using a^2 + b^2 = 1989 - c^2 for c = 1 thru 44.

Nothing is presenting itself as an easier approach than trial and error for the third one.

First two were obvious if you know your right triangles.

[Guest]

Ya same problem here.

[curr3nt]

Wrote a quick program and found 11 unique sets.

edit - Interesting that only one number is odd in each set. None of them are three odd numbers.

Edited February 9, 2011 by curr3nt

[curr3nt]

For uniqueness lets say: a <= b <= c

We know the largest value c could be is 44. Because the square root of 1989 is ~= 44.

We can determine a lowest value using the same value for each variable.

c^2 + c^2 + c^2 = 1989 => 3c^2 = 1989 => c^2 = 663 => c ~= 25

So 25 < c <=44 for each set that could exist.

This cuts down on trial and error attempts.

edit - changed 25 <= c to 25 < c

Edited February 9, 2011 by curr3nt

[Guest]

Cuts down, doesn't eliminate.

[curr3nt]

Let x = 1989 - c^2

c = 26 then x = 1313

...

c = 44 then x = 53

Using a^2 + b^2 = x and a <= b <= c

We can limit the range of b for each x. sqrt(x/2) <= b <= sqrt(x) { replace sqrt(x) with c if sqrt(x) > c since b <= c }

So for the 19 cases of c, I come up with 113 possible values of b and c for trial and error.

Examples: (The first comparison is a < not a <= because sqrt(x/2) is not an integer for any of the values of c)

c = 26 => 25 < b <= 26(36) { replaced with c } => 1 possible value

c = 31 => 22 < b <= 31(32) { replaced with c } => 9 possible values

c = 32 => 21 < b <= 31 => 10 possible values

c = 33 => 21 < b <= 30 => 9 possible values

c = 44 => 5 < b <= 7 => 2 possible values

Edited February 9, 2011 by curr3nt

[curr3nt]

Cuts down, doesn't eliminate.

Eliminating sets to 113 to test is better than just plugging in numbers. 44 possible values in each variable 44 * 44 * 44 = 85184 sets.

Makes the trial and error approach more managable.

[Thalia]

What is the exact text of the problem?

[Guest]

What is the exact text of the problem?

Thank you, Thalia, Darth Legion and Curr3nt for the great suggestions. I need to go over them slowly and really digest them if I am to understand what to do. Thalia, here is the exact wording of the problem: "Find three ways to express 1989 as the sum of the squares of three positive integers." My trial and error strategy does not work and it takes me a very long time to a path of nowhere.

[Thalia]

I made a table of squares. Starting from the bottom (44^2), I subtracted and looked for two squares that would fit the remainder. 44^2=1936. 1989-1936=53. 53=49+4. 44, 7, 2. I found that going down by twos worked. 42^2, 40^2, 38^2, etc. 42^2=1764. 1989-1764=225. Find two sqares that equal 225.

1 1

2 4

3 9

4 16

5 25

6 36

7 49

8 64

9 81

10 100

11 121

12 144

13 169

14 196

15 225

16 256

17 289

18 324

19 361

20 400

21 441

22 484

23 529

24 576

25 625

26 676

27 729

28 784

29 841

30 900

31 961

32 1024

33 1089

34 1156

35 1225

36 1296

37 1369

38 1444

39 1521

40 1600

41 1681

42 1764

43 1849

44 1936

[Guest]

I made a table of squares. Starting from the bottom (44^2), I subtracted and looked for two squares that would fit the remainder. 44^2=1936. 1989-1936=53. 53=49+4. 44, 7, 2. I found that going down by twos worked. 42^2, 40^2, 38^2, etc. 42^2=1764. 1989-1764=225. Find two sqares that equal 225.

1 1

2 4

3 9

4 16

5 25

6 36

7 49

8 64

9 81

10 100

11 121

12 144

13 169

14 196

15 225

16 256

17 289

18 324

19 361

20 400

21 441

22 484

23 529

24 576

25 625

26 676

27 729

28 784

29 841

30 900

31 961

32 1024

33 1089

34 1156

35 1225

36 1296

37 1369

38 1444

39 1521

40 1600

41 1681

42 1764

43 1849

44 1936

Thanks, Thalia, for your hardwork and for showing me the way to the light. I feel so blessed to have met you.

[curr3nt]

Well, I guess it is ok to give you more info...

Using the limits of c here are the limits of b.

If c = 26 then 25.62 <= b <= 36.24 --- 26 <= b <= 26

If c = 27 then 25.10 <= b <= 35.50 --- 26 <= b <= 27

If c = 28 then 24.55 <= b <= 34.71 --- 25 <= b <= 28

If c = 29 then 23.96 <= b <= 33.88 --- 24 <= b <= 29

If c = 30 then 23.33 <= b <= 33.00 --- 24 <= b <= 30

If c = 31 then 22.67 <= b <= 32.06 --- 23 <= b <= 31

If c = 32 then 21.97 <= b <= 31.06 --- 22 <= b <= 31

If c = 33 then 21.21 <= b <= 30.00 --- 22 <= b <= 30

If c = 34 then 20.41 <= b <= 28.86 --- 21 <= b <= 28

If c = 35 then 19.54 <= b <= 27.64 --- 20 <= b <= 27

If c = 36 then 18.61 <= b <= 26.32 --- 19 <= b <= 26

If c = 37 then 17.61 <= b <= 24.90 --- 18 <= b <= 24

If c = 38 then 16.51 <= b <= 23.35 --- 17 <= b <= 23

If c = 39 then 15.30 <= b <= 21.63 --- 16 <= b <= 21

If c = 40 then 13.95 <= b <= 19.72 --- 14 <= b <= 19

If c = 41 then 12.41 <= b <= 17.55 --- 13 <= b <= 17

If c = 42 then 10.61 <= b <= 15.00 --- 11 <= b <= 15

If c = 43 then  8.37 <= b <= 11.83 ---  9 <= b <= 11

If c = 44 then  5.15 <= b <=  7.28 ---  6 <= b <=  7

From that here are the 113 values of b and c you would have to check for integer values of a. Using a spreedsheet tool like excel or google docs it is easy to find the 11 unique sets of [a,b,c].

[?, b, c] : 1989 -  c^2 -  b^2 = a^2

[?,26,26] : 1989 - 26^2 - 26^2 = a^2

[?,26,27] : 1989 - 27^2 - 26^2 = a^2

[?,27,27] : 1989 - 27^2 - 27^2 = a^2

[?,25,28] : 1989 - 28^2 - 25^2 = a^2

[?,26,28] : 1989 - 28^2 - 26^2 = a^2

[?,27,28] : 1989 - 28^2 - 27^2 = a^2

[?,28,28] : 1989 - 28^2 - 28^2 = a^2

[?,24,29] : 1989 - 29^2 - 24^2 = a^2

[?,25,29] : 1989 - 29^2 - 25^2 = a^2

[?,26,29] : 1989 - 29^2 - 26^2 = a^2

[?,27,29] : 1989 - 29^2 - 27^2 = a^2

[?,28,29] : 1989 - 29^2 - 28^2 = a^2

[?,29,29] : 1989 - 29^2 - 29^2 = a^2

[?,24,30] : 1989 - 30^2 - 24^2 = a^2

[?,25,30] : 1989 - 30^2 - 25^2 = a^2

[?,26,30] : 1989 - 30^2 - 26^2 = a^2

[?,27,30] : 1989 - 30^2 - 27^2 = a^2

[?,28,30] : 1989 - 30^2 - 28^2 = a^2

[?,29,30] : 1989 - 30^2 - 29^2 = a^2

[?,30,30] : 1989 - 30^2 - 30^2 = a^2

[?,23,31] : 1989 - 31^2 - 23^2 = a^2

[?,24,31] : 1989 - 31^2 - 24^2 = a^2

[?,25,31] : 1989 - 31^2 - 25^2 = a^2

[?,26,31] : 1989 - 31^2 - 26^2 = a^2

[?,27,31] : 1989 - 31^2 - 27^2 = a^2

[?,28,31] : 1989 - 31^2 - 28^2 = a^2

[?,29,31] : 1989 - 31^2 - 29^2 = a^2

[?,30,31] : 1989 - 31^2 - 30^2 = a^2

[?,31,31] : 1989 - 31^2 - 31^2 = a^2

[?,22,32] : 1989 - 32^2 - 22^2 = a^2

[?,23,32] : 1989 - 32^2 - 23^2 = a^2

[?,24,32] : 1989 - 32^2 - 24^2 = a^2

[?,25,32] : 1989 - 32^2 - 25^2 = a^2

[?,26,32] : 1989 - 32^2 - 26^2 = a^2

[?,27,32] : 1989 - 32^2 - 27^2 = a^2

[?,28,32] : 1989 - 32^2 - 28^2 = a^2

[?,29,32] : 1989 - 32^2 - 29^2 = a^2

[?,30,32] : 1989 - 32^2 - 30^2 = a^2

[?,31,32] : 1989 - 32^2 - 31^2 = a^2

[?,22,33] : 1989 - 33^2 - 22^2 = a^2

[?,23,33] : 1989 - 33^2 - 23^2 = a^2

[?,24,33] : 1989 - 33^2 - 24^2 = a^2

[?,25,33] : 1989 - 33^2 - 25^2 = a^2

[?,26,33] : 1989 - 33^2 - 26^2 = a^2

[?,27,33] : 1989 - 33^2 - 27^2 = a^2

[?,28,33] : 1989 - 33^2 - 28^2 = a^2

[?,29,33] : 1989 - 33^2 - 29^2 = a^2

[?,30,33] : 1989 - 33^2 - 30^2 = a^2

[?,21,34] : 1989 - 34^2 - 21^2 = a^2

[?,22,34] : 1989 - 34^2 - 22^2 = a^2

[?,23,34] : 1989 - 34^2 - 23^2 = a^2

[?,24,34] : 1989 - 34^2 - 24^2 = a^2

[?,25,34] : 1989 - 34^2 - 25^2 = a^2

[?,26,34] : 1989 - 34^2 - 26^2 = a^2

[?,27,34] : 1989 - 34^2 - 27^2 = a^2

[?,28,34] : 1989 - 34^2 - 28^2 = a^2

[?,20,35] : 1989 - 35^2 - 20^2 = a^2

[?,21,35] : 1989 - 35^2 - 21^2 = a^2

[?,22,35] : 1989 - 35^2 - 22^2 = a^2

[?,23,35] : 1989 - 35^2 - 23^2 = a^2

[?,24,35] : 1989 - 35^2 - 24^2 = a^2

[?,25,35] : 1989 - 35^2 - 25^2 = a^2

[?,26,35] : 1989 - 35^2 - 26^2 = a^2

[?,27,35] : 1989 - 35^2 - 27^2 = a^2

[?,19,36] : 1989 - 36^2 - 19^2 = a^2

[?,20,36] : 1989 - 36^2 - 20^2 = a^2

[?,21,36] : 1989 - 36^2 - 21^2 = a^2

[?,22,36] : 1989 - 36^2 - 22^2 = a^2

[?,23,36] : 1989 - 36^2 - 23^2 = a^2

[?,24,36] : 1989 - 36^2 - 24^2 = a^2

[?,25,36] : 1989 - 36^2 - 25^2 = a^2

[?,26,36] : 1989 - 36^2 - 26^2 = a^2

[?,18,37] : 1989 - 37^2 - 18^2 = a^2

[?,19,37] : 1989 - 37^2 - 19^2 = a^2

[?,20,37] : 1989 - 37^2 - 20^2 = a^2

[?,21,37] : 1989 - 37^2 - 21^2 = a^2

[?,22,37] : 1989 - 37^2 - 22^2 = a^2

[?,23,37] : 1989 - 37^2 - 23^2 = a^2

[?,24,37] : 1989 - 37^2 - 24^2 = a^2

[?,17,38] : 1989 - 38^2 - 17^2 = a^2

[?,18,38] : 1989 - 38^2 - 18^2 = a^2

[?,19,38] : 1989 - 38^2 - 19^2 = a^2

[?,20,38] : 1989 - 38^2 - 20^2 = a^2

[?,21,38] : 1989 - 38^2 - 21^2 = a^2

[?,22,38] : 1989 - 38^2 - 22^2 = a^2

[?,23,38] : 1989 - 38^2 - 23^2 = a^2

[?,16,39] : 1989 - 39^2 - 16^2 = a^2

[?,17,39] : 1989 - 39^2 - 17^2 = a^2

[?,18,39] : 1989 - 39^2 - 18^2 = a^2

[?,19,39] : 1989 - 39^2 - 19^2 = a^2

[?,20,39] : 1989 - 39^2 - 20^2 = a^2

[?,21,39] : 1989 - 39^2 - 21^2 = a^2

[?,14,40] : 1989 - 40^2 - 14^2 = a^2

[?,15,40] : 1989 - 40^2 - 15^2 = a^2

[?,16,40] : 1989 - 40^2 - 16^2 = a^2

[?,17,40] : 1989 - 40^2 - 17^2 = a^2

[?,18,40] : 1989 - 40^2 - 18^2 = a^2

[?,19,40] : 1989 - 40^2 - 19^2 = a^2

[?,13,41] : 1989 - 41^2 - 13^2 = a^2

[?,14,41] : 1989 - 41^2 - 14^2 = a^2

[?,15,41] : 1989 - 41^2 - 15^2 = a^2

[?,16,41] : 1989 - 41^2 - 16^2 = a^2

[?,17,41] : 1989 - 41^2 - 17^2 = a^2

[?,11,42] : 1989 - 42^2 - 11^2 = a^2

[?,12,42] : 1989 - 42^2 - 12^2 = a^2

[?,13,42] : 1989 - 42^2 - 13^2 = a^2

[?,14,42] : 1989 - 42^2 - 14^2 = a^2

[?,15,42] : 1989 - 42^2 - 15^2 = a^2

[?, 9,43] : 1989 - 43^2 -  9^2 = a^2

[?,10,43] : 1989 - 43^2 - 10^2 = a^2

[?,11,43] : 1989 - 43^2 - 11^2 = a^2

[?, 6,44] : 1989 - 44^2 -  6^2 = a^2

[?, 7,44] : 1989 - 44^2 -  7^2 = a^2

edit - fixing spacing.

Edited February 10, 2011 by curr3nt

[Guest]

Well, I guess it is ok to give you more info...Using the limits of c here are the limits of b. From that here are the 113 values of b and c you would have to check for integer values of a. Using a spreedsheet tool like excel or google docs it is easy to find the 11 unique sets of [a,b,c].

I am really impressed at your knowledge and expertise, Curr3nt. You are a true math puzzle solver.

[Guest]

a^2 + b^2 + c^2 = 1989

9^2 + 12^2 + 42^2 = 1989