Friendly Triples

We know that a friendly triple has $a = bc, b = ac$ , and $c = ab$ . Multiplying these three equations together, we have $abc = (abc)^2$ . Solving this as a quadratic in terms of $(abc)$ , we have $abc = 0$ or $abc = 1$ .

For the $abc = 0$ case, we know that one of $a, b$ or $c$ is 0. But from the equations, the rest must also be 0. So this gives us one triple, (0, 0, 0).

For the $abc=1$ case, we can safely [Why safely? - Calvin] say that $a = 1/(bc)$ , $b = 1/(ac)$ and $c = 1/(ab)$ . Multiplying with the original system of three equations, we know that $a^2 = 1$ , $b^2 = 1$ and $c^2 = 1$ , so $a, b$ and $c$ are either 1 or -1. But since the product $abc$ is positive, either zero or two of $a, b$ , and $c$ must be negative. So, we have four more triples, (1, 1, 1), and the three permutations of (-1, -1, 1), for a total of five triples.

Note that ab=c, bc=a and ca=b. Multiplying the three equations, (abc)^2=abc, abc=0 or abc=1. When abc=0, wlog a=0, comparing the three equations, b=c=0, therefore we get a=b=c=0. When abc= 1, wlog divide both sides by ab=c, we get c^2=1, c=1 or c=-1. Similarly, we get a=1 or a=-1, b=1or b=-1. By checking all possibilities, the solutions are (a,b,c)=(1,1,1),(-1,-1,1),(-1,1,-1),(1,-1,-1). Therefore, there are a total of 5 solutions.

From the following information, we know that:

(1) $a=bc$

(2) $b=ac$

(3) $c=ab$

Using (1) and (2),

$b=bc^2$

$b(c^2-1)=0$

Thus, $b=0$ or $c^2-1=0\Rightarrow c=\pm 1$ .

If $b=0$ , from (1) and (2), $a=c=0$ . Thus, $(a,b,c)=(0,0,0)$ .

If $c=1$ , from (1) or (2), $a=b$ . Since $1=c=ab=a^2$ , $a=b=\pm 1$ . Thus, $(a,b,c)=(1,1,1)$ or $(a,b,c)=(-1,-1,1)$ .

Take note that the we can also take all the possible permutations of $(a,b,c)=(-1,-1,1)$ since the equations are symmetric. There are $\binom{3}{2}$ such permutations.

If $c=-1$ , from (1) or (2), $a=-b$ . Since $-1=c=-a^2$ , $a=b=\pm 1$ . Thus, $(a,b,c)=(\pm1,\mp1,-1)$ . Notice, however that this is just a permutation of the cases above.

Thus, there are five ( $5$ ) solutions, namely $(0,0,0)$ , $(1,1,1$ , and all the permutations of $(-1,-1,1)$ .

We have $a = bc, b = ac, c = ab$ . Multiplying all 3 equations together, we get $abc = a^2 b^2 c^2$ , so $0 = a^2b^2c^2-abc = abc(abc-1)$ . Hence, we have the following cases:

Case 1. $abc=0$ . If $a=0$ , then $b=ac = 0$ and $c=ab=0$ , so the only solution is $(0,0,0)$ .

Case 2. $abc = 1$ . Then, $a^2 = a \times a = a \times bc = 1$ , so $a = \pm 1$ . Similarly, $b = \pm 1, c= \pm 1$ . Since $abc=1$ , an even number of these terms must be $-1$ . We have the possibilities $(1, 1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1)$ , and it is clear that all of these satisfy the condition, so there are $4$ solutions.

Hence, in total, there are $5$ solutions.

You know that: 0 x 0 = 0 -0 x 0 = 0 1 x 1 = 1 -1 x -1 = 1

Because 0 and -0 are the same, there are 3 possibilities. So first, one possibility of an ordered triple is (0,0,0). Then, you know (1,1,1), the hard part is the -1s.

But, if you do (-1,-1,1) it will work since -1 x 1 = -1, and -1 x -1 = 1.

Then, you do all the other permutations (there are two more).

SO the answer is 5 since there is (0,0,0), (1,1,1), (-1,-1,1), (-1,1,-1) and (1,-1,-1)

We can see that (1,1,1) and (0,0,0) work. We can also see that only 1's and -1's will work, so the only other solution is (-1,-1,1), which there are 3 permutations of.