Question 1208809: How many integer solutions (a, b, c) make the equation a2+b2+c2 = 169 true?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!
For any solution we find, changing the sign of any one of the numbers gives another solution, because (-a)^2 is equal to a^2. So to start with we only need to look for non-negative solutions.
The problem does not specify positive integers, so 0 is a possibility. And since 169 = 13^2, the triple of numbers (13,0,0) is a solution.
Those three numbers can be in any order -- i.e., (0,13,0) and (0,0,13) are also solutions. So there are 3 solutions using the non-negative integers 13, 0, and 0.
But any of those numbers can be replace by its opposite. The "opposite" of 0 is still 0, so we can't change the sign of any 0 to get a new solution. But we can replace each 13 with -13 to get a new solution.
Summary of the problem so far:
13, 0, and 0 satisfy the equation (1 solution);
there are 3 ways to arrange those three numbers (1*3 = 3 solutions);
we can change the sign of the 13 (3*2 = 6 solutions).
So there are 6 solutions using the number 13, 0, 0, and their opposites.
Now look for other solutions. To do that, use a "greedy" algorithm -- that is, try the largest numbers first. We have used 13, so now look at 12.
12^2 = 144; 169-144 = 25; and that can be written as 16+9 = 4^2+3^2.
So the numbers 12, 4, and 3 provide another solution -- or, in fact, a large set of solutions.
Rearranging the three numbers and changing any of their signs, similar to what we did with the three number 13, 0, and 0, we get the following summary:
12, 4, and 3 satisfy the equation (1 solution);
there are 6 ways to arrange those three numbers (1*6 = 6 solutions);
we can change the sign, or not change the sign, of EACH of the three numbers (6*2*2*2 = 48 solutions).
Continuing with our greedy algorithm, we quickly find that there is no other set of three numbers which will give a solution to the given equation.
So...
ANSWER: There are 6+48 = 54 solutions to the given equation
----------------------------------------------------------
Thanks to tutor @ikleyn for pointing out the missing solutions in my original answer.
I used 0 in the set 13, 0, and 0; so I should have remembered to include the set 12, 5, and 0.
The number of missing solutions in my original solution is...
12, 5, 0: (1 solution)
arranging those in any of 6 ways: 1*6 = 6 solutions
changing the sign of either or both of the 12 and 5: 6*2*2 = 24 solutions
Final (corrected) answer: 54+24 = 78
Answer by ikleyn(52747)  (Show Source):
You can put this solution on YOUR website! .
How many integer solutions (a, b, c) make the equation a2+b2+c2 = 169 true?
~~~~~~~~~~~~~~~~~~~~~~~~
To the basic triples (13,0,0), (12,4,3), found by @greenestamps, the triple (12,5,0) should be added.
With permutations, it produces 6 solutions with non-negative integer numbers.
Playing with the signs, it gives 6*2*2 = 24 solutions to be added to 54 solutions, counted by @greenestamps.
In all, it gives 54 + 24 = 78 solutions to the given equation in integer numbers.
ANSWER. There are 78 solutions to the given equation in integer numbers.
|
|
|
