Triples of Integers

We have $288a+b^2c^2=288c+a^2b^2\Rightarrow 288(c-a)=b^2(c-a)(c+a)$ and since that $c>a$ we have $c-a>0$ and by the last equal,we can conclude that

$(*)\quad 288=b^2(a+c)$

Since that $288=2^53^2$ and $b^2|288$ , we have $b=2^x3^y$ , where $x\le 2, y\le 1\Rightarrow b|12$ .

Now, we are going divide in three cases:

$(1)\quad b=c$

By $(*)$ , we have $a<c\Rightarrow a+c<2c\Rightarrow 288=b^2(a+c)<2b^2c=2b^3$

and soo $b>5$ . How $b|12$ , then $b=6, 12$ .

If $b=c=6, b=c=12$ , by the $(*)$ we have $a=2, -10$ , respectively, and in both cases, we have $a<c, a\le b\le c$ . Solutions $(a, b, c)=(2, 6, 6); (-10, 12, 12)$ .

$(2)\quad b=a$

We have $a<c\Rightarrow 2a<a+c\Rightarrow 2b^3=2ab^2<b^2(a+c)=288$ , where $b<5$ and $b|12$ , then $(**)\quad b=-12, -6, -4, -3, -2, -1, 1, 2, 3, 4$ .

If $b<0$ , we have $0<288=b^2(a+c)=b^2(b+c), b^2>0$ , where $b+c>0\Rightarrow c>-b>0>b=a$ , and soo every negative value for $b$ in $(**)$ generates a solution. Explicity, we have $c=2(\frac{12}{b})^2-b$

So we have $6$ solutions.

By test for error, we conclude that $b=1, 2, 3, 4$ generates solutions which satisfy $a\le b\le c, a<c$ and we have $4$ new solutions.

$(3)\quad a<b<c$

By a similar simplification to $(*)$ we have $(***)\quad a^2(b+c)=b^2(a+c)=c^2(a+b)=288$ .

By the same argument, we have $c|12$ and $a+b<2c\Rightarrow 288=c^2(a+b)<2c^3, c>5$ and we have $c=6, 12$ .

If $c=6$ , $(***)$ give us $a+b=\frac{288}{6^2}=8, a^2b+6a^2=ab^2+6b^2\Rightarrow ab=-6(a+b)=-48$ and we have $a=-4, b=12$ since that $a<b$ . It's a contradiction with $b<c$ .

If $c=12$ $(***)$ give us $a+b=\frac{288}{12^2}=2, a^2b+12a^2=ab^2+12b^2\Rightarrow ab=-12(a+b)=-24$ and we have $a=-4, b=6$ since that $a<b$ . Actually, $(a, b, c)=(-4, 6, 12)$ is a solution.

So we have $2+6+4+0+1=13$ solutions.

We have three equalities: $288a+b^2c^2=288b+c^2a^2$ $288b+c^2a^2=288c+a^2b^2$ $288a+b^2c^2=288c+a^2b^2$ We can manipulate these equalities. $288(a-b)+c^2(b^2-a^2)=0$ $288(a-b)+c^2(b-a)(b+a)=0$ And now, we divide by $a-b$ , $288=c^2(a+b)\text{ or }a=b\quad (1)$ Similarly, we have: $288=b^2(a+c)\text{ or }a=c\quad (2)$ $288=a^2(b+c)\text{ or }b=c\quad (3)$ However, we are given $c>a$ , so the left part of (2) is always true. We will focus on that one. We know that $b^2$ divides 288, and so we have 12 cases.

For each $b^2$ , we can try to find $a,c$ such that $b=a$ or $b=c$ , since that eliminates an equation.

Say $a=b$ $c=\dfrac{288}{b^2}-b$ Now, (1) and (2) are satisfied. (2) implies (3) since $a=b$ , and so our assignments satisfy the system.

Similarly, $b=c$ $a=\dfrac{288}{b^2}-b$ However, we must confirm that $a<c$ in each solution.

$b=1$ , $a+c=288$ --> $(1,1,287)$ is a solution ( $a=b$ )

$b=2$ , $a+c=72$ --> $(2,2,70)$ is a solution ( $a=b$ )

$b=3$ , $a+c=32$ --> $(3,3,29)$ is a solution ( $a=b$ )

$b=4$ , $a+c=18$ --> $(4,4,14)$ is a solution ( $a=b$ )

$b=6$ , $a+c=8$ --> $(2,6,6)$ is a solution ( $b=c$ )

$b=12$ , $a+c=2$ --> $(-10,12,12)$ is a solution ( $b=c$ )

$b=-1$ , $a+c=288$ --> $(-1,-1,289)$ is a solution ( $a=b$ )

$b=-2$ , $a+c=72$ --> $(-2,-2,74)$ is a solution ( $a=b$ )

$b=-3$ , $a+c=32$ --> $(-3,-3,35)$ is a solution ( $a=b$ )

$b=-4$ , $a+c=18$ --> $(-4,-4,22)$ is a solution ( $a=b$ )

$b=-6$ , $a+c=8$ --> $(-6,-6,14)$ is a solution ( $a=b$ )

$b=-12$ , $a+c=2$ --> $(-12,-12,14)$ is a solution ( $a=b$ )

So far, we have 12 solutions. Now, if more solutions exist, then $a\neq b\neq c$ . Therefore, $a^2(b+c)=b^2(a+c)=c^2(a+b)=288$ If this works, $a^2|288$ , $b^2|288$ , and $c^2|288$ . Also, $b^2(a+c)=288$ . This means that $a,b,c>3$ , since if one is below, then the sum of the other two must be at least $32>12+12=24$ . Now, we have 6 possible values, and it is easy to confirm that the only other solution is $(-4,6,12)$ .

The answer is $\boxed{13}$ .

Let's drop the condition $a \le b \le c$ first and consider all solutions where not all a, b, c are equal.

Consider the case where two of the variables are equal, i.e. a=b without loss of generality. Then:

$288a + b^2 c^2 = 288c + a^2 b^2 \implies 288(a-c) = b^2(a^2 - c^2) = b^2(a-c)(a+c)$

and since a, b, c are not all equal, we have $a\ne c$ so $288 = b^2(a+c) = b^2(b+c)$ . Thus, the set of solutions corresponds to each integer b such that $b^2 | 288$ , since we can just let a=b and $c = \frac{288}{b^2} - c$ . This gives 12 solutions:

$b=\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12.$

The remaining case is where a, b, c are all distinct. Following the above reasoning, we get:

$288 = a^2(b+c) = b^2(a+c) = c^2(a+b)$

so $a,b,c = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ . Since $-24 \le b+c \le 24$ , we can't have $a = \pm 1, \pm 2, \pm 3$ since that would imply $b+c = 288, 72, 32$ respectively. Thus, $a, b, c = \pm 4, \pm 6, \pm 12$ .

Consider the case where a=4. This gives $b+c = 288/16 = 18$ so we must have (b,c) = (6,12) or (12,6) which contradicts $b^2(a+c) = 288$ . The case a=-4 also gives b+c=18, but here, this works so we get a solution of (-4, 6, 12).

Excluding the case of -4 and +4, we're left with $a, b, c = \pm 6, \pm 12$ . This is clearly impossible since a, b, c are all multiples of 6 and that would give $216 | b^2(a+c)$ .

Conclusion: 13 solutions, 12 of which corresponds to a=b or b=c and the final one is (-4, 6, 12).

Since $a<c$ , we have $288(a-c)=b^2(a^2-c^2)\implies 288 = b^2(a+c)$ .

If $a < b < c$ , we similarly have $288=c^2(a+b)$ and $288=a^2(b+c) > a^2(2a)= 2a^3$ . Since $a^2|288$ , it follows that $a\le 4$ . Then $b+c=\frac{288}{a^2}\ge\frac{288}{16}= 18$ . However, note that $b,c$ must equal one of the folllowing: $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ with $b<c$ . Thus, the only way to satisfy $b+c\ge18$ is when $a=\pm4,b=6,c=12$ . Checking, only $a=-4,b=6,c=12$ works.

If $a=b < c$ , $288=b^2(a+c)$ reduces to $288=a^2(a+c) > a^2(2a)=2a^3$ , so as before, $a\le 4$ . Thus, out of the possible values: $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ , only $a=\pm1,\pm2,\pm3,\pm4,-6,-12$ work, for a total of $10$ solutions.

If $a<b=c$ , we have $288=c^2(a+c) < c^2(2c)=2c^3$ , so $c\ge 6$ . Again, out of the possible values: $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ , only $c=6,12$ work, for a total of $2$ solutions.

As a result, there are $1+10+2=\boxed{13}$ solutions.

Note that $(0,0,0)$ is not a solution since $a<c$ . Since $a<c$ , then if any two variables are equal we can either have $a=b$ or $b=c$ . But, before that we will try to find integer solutions such that $a,b,c$ are distinct.

Solving the given equations will give us the following system of equations:

$a^2(b+c)=288$

$b^2(c+a)=288$

$c^2(a+b)=288$ Now we note that since $a,b,c$ are integers, then $a^2,b^2,c^2$ have to be square numbers that divide $288$ Also, $(b+c),(a+b),(a+c)|288$ [Divides]. So, $a,b,c \in ( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 )$ . Clearly, $-12 \leq a,b,c \leq 12$ and, $0\leq (a+c),(b+c),(a+b) \leq 24$ [Since they can only be positive].

So, $a,b,c \in { \pm 4, \pm 6, \pm 12 }$ Suppose, if $a=4$ then since the integers need to be distinct and our conditions satisfied the only possible triple is $(4,6,12)$ . Since $4+6=10$ which does not divide $288$ , this triple is not a solution.

If, $a=-4$ then $b=6$ and it can’t equal $4$ or $12$ . It can’t equal $4$ because then we have $c=0$ which is not possible. It can’t equal $12$ because we look for distinct triples. So, we have $(-4,6,12)$ as a possible candidate. We check that $(-4,6,12)$ does, indeed, satisfy the given equations. Now if $a=-6$ , we can have $b=4$ and, $c=12$ . Note that $a+b=-2$ , but since $c^2$ is obviously a positive factor of $288$ so must be $a+b$ . So, we reject this case as well. An analogous argument for $a=-12$ tell us that there can be no solutions.

Now, when $a=b$ :

Note that in this case \(b \neq c\( Our original equations reduce to:

\(288b + b^2 c^2 = 288b + c^2 b^2 = 288c + b^4 \)

$\Rightarrow 288b + c^2 b^2 = 288c + b^4$ $288(b-c) = b^2( b^2 - c^2)$ $288=b^2(b+c)$ [Since $b \neq c$ ] $b \in ( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 )$

Clearly there will exist integer solutions to $c$ for all the above $b$ . But, we have to consider the order such that $a < c$ . But, we see that if we ignore the condition we have 12 triples such that $a=b$ .

For example if there exists such triple in the above triples such that $a>c$ , then we can simply exchange the value of $a$ with $c$ and our condition becomes $b=c$ .We can do this, since we see that if we replace the relative position $a,b,c$ in the original equations, the equations remain unchanged.

Let 288 + b^2 c^2 = 288b + c^2 a^2 = 288c + a^2 b^2=k\ for some integer \(k \neq 0

$288(a+b+c) + b^2 c^2 + c^2 a^2 + a^2 b^2 -3k=0$

Let $f(a,b,c)= 288(a+b+c) + b^2 c^2 + c^2 a^2 + a^2 b^2 -3k$

Clearly this is the same as $f(a,c,b)$ , $f(c,a,b)$ , $f(c,b,a)$ . This concludes our proof that there are 12 solutions to the case when either $a=b$ or $b=c$ [ $a \neq c$ ] We don’t need to check separately for $b=c$ since we have already counted those triples. Total number of solutions is $\boxed {13}$

divides in 3 case case 1 a=b<c equation is a^2(a+c)=288=2^5x3^2=2x(2^2)^2x3^2 because in LHS a^2, a=+-1,+-2,+-3,+-4,+-6,+-12 a=b=-1, c=289 a=b=1, c=287 a=b=-2, c=74 a=b=2, c=70 a=b=-3, c=35 a=b=3, c=29 a=b=-4, c=22 a=b=4, c=14 a=b=-6, c=14 a=b=6, c=2 (a>c) (not solution) a=b=-12, c=14 a=b=12, c=-10, notsolution so there are 10 solution for first case Case 2 a<b=c the equation is b^2(a+b)=288 similiarly only b=c=12, a=-10 and b=c=6, a=2 Case 3 a<b<c only a=-4,b=6, c=12

So the total solution is 13

Let's drop the condition a≤b≤c first and consider all solutions where not all a, b, c are equal.Consider the case where two of the variables are equal, i.e. a=b without loss of generality. Then:288a+b2c2=288c+a2b2⟹288(a−c)=b2(a2−c2)=b2(a−c)(a+c)and since a, b, c are not all equal, we have a≠c so 288=b2(a+c)=b2(b+c). Thus, the set of solutions corresponds to each integer b such that b2|288, since we can just let a=b and c=288b2−c. This gives 12 solutions:b=±1,±2,±3,±4,±6,±12.The remaining case is where a, b, c are all distinct. Following the above reasoning, we get:288=a2(b+c)=b2(a+c)=c2(a+b)so a,b,c=±1,±2,±3,±4,±6,±12. Since −24≤b+c≤24, we can't have a=±1,±2,±3 since that would imply b+c=288,72,32 respectively. Thus, a,b,c=±4,±6,±12.Consider the case where a=4. This gives b+c=288/16=18 so we must have (b,c) = (6,12) or (12,6) which contradicts b2(a+c)=288. The case a=-4 also gives b+c=18, but here, this works so we get a solution of (-4, 6, 12).Excluding the case of -4 and +4, we're left with a,b,c=±6,±12. This is clearly impossible since a, b, c are all multiples of 6 and that would give 216|b2(a+c).Conclusion: 13 solutions, 12 of which corresponds to a=b or b=c and the final one is (-4, 6, 12).

We split the problem into three cases:

Case 1 , $\textbf{a<b<c}$ : In this case, we know that none of $a-b,b-c,c-a$ are $0$ , and so we can divide through by them. $288a + b^2c^2 = 288b + c^2a^2 \to 288(a-b) = c^2(a^2-b^2) \to$ $288 = c^2(a+b)$ , and through similar algebra, we arrive at $288 = a^2(b+c) = b^2(c+a) = c^2(a+b)$ . Because $a,b,c \in \mathbb{Z}, a^2,b^2,c^2\mid288$ (otherwise, the sum of two integers is not an integer, which is impossible). Thus, $a^2,b^2,c^2 \mid 3^2\cdot2^5 \to$ $a,b,c \in \{\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\}$ . We proceed to test each possible value of $a^2$ by plugging it into the equation $288 = a^2(b+c)$ : since $b+c\leq18$ (remember, $b < c$ ), $a^2\geq16 \to a \in \{\pm4,\pm6,\pm12\}$ . We can verify that $a = -4$ yields $(a,b,c) = (-4,6,12)$ , which also works in the other two equalities. $a = 6,12$ yield no solutions because $a<b<c$ , and then the solutions we get from $a = -6,-12$ do not work in the other equalities. Therefore, this case yields only the ordered triple $(-4,6,12)$ .

Case 2 , $\textbf{a = b < c}$ : In this case, we can substitute all the $b$ 's with $a$ 's to get the equality $288a + a^2c^2 = 288c + a^2a^2 = 288c + a^4 \to$ $288(a-c) = a^2(a^2-c^2) = a^2(a+c)(a-c)$ . Because $a<c$ , we can divide through by $a-c$ to get $288 = a^2(a+c) \to \frac{288-a^3}{a^2} = c$ . Clearly, $a^2 \mid a^3$ , so we need $a^2 \mid 288$ , and we already know that the values of $a$ that satisfy this are $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ . This gives us $12$ ordered triples of the form $(a,a,\frac{288-a^3}{a^2})$ . However, we must remember that $a < c \to a < \frac{288-a^3}{a^2} \to 2a^3 < 288 \to a < \sqrt[3]{144}$ . Thus, we have $10$ ordered triples in this case, because we must remove $(6,6,3)$ and $(12,12,-10)$ .

Case 3 , $\textbf{a < b = c}$ : In this case, we use manipulations very similar to those in Case 2 to get that $288 = c^2(c+a)$ and $a < c$ . Notice that replacing the $a$ 's with $c$ 's and vice-versa yields $288 = a^2(a+c)$ and $a > c$ , which gives precisely the two solutions we eliminated from Case 2 with just the value for $a$ switched with the value for $c$ . Thus, we get two solutions in this case, namely $(a,b,c) = (3,6,6), (-10,12,12)$ .

We can verify trivially that there are no repeats in these three cases because in each case, the behavior of $b$ is different. In each of Case 2's solutions, for example, $b = a$ , and so none of those solutions can also be a solution in Case 1 or Case 3.

Therefore, we have a total of $1 + 10 + 2 = \fbox{13}$ ordered triples of integers $(a,b,c)$ which satisfy the given conditions.

Let's consider the cases of $a < b$ and $a \le b$ separately. If $a < b$ then none of $a-b$ , $a-c$ or $b-c$ are zero so we can take the differences between each of the three equalities in the question and divide through by the above factors to obtain $a^2(b+c) = 288, b^2(a+c) = 288$ and $c^2(a+b) = 288$ . We therefore have $a^2,b^2, c^2 \mid 288 = 2^5 \times 3^2$ so $a,b,c \mid 2^2 \times 3 = 12$ . If $|a| \le 3$ then there is no way that $b+c = 288/a^2 \ge 32$ if $b,c \mid 12$ so we have $a \in \{-12,-6,-4,4,6,12\}$ . Since $a<b<c$ , the value of $a$ cannot be 6 or 12 so this reduces the options to $a \in \{-12,-6,-4,4\}$ . Also since $0<a+b$ we cannot have $a \in \{-12,-6\}$ . So we need only examine $a = \pm 4$ . We need $b+c = 18$ and the only possibility for this is $b = 6$ and $c=12$ , and we find that the only valid solution here is $(a,b,c) = (-4,6,12)$ .

Now we deal with the case $a=b$ . We now only have one equality to examine and this reduces to $a^2(a+c) = 288$ . By the same logic as before we have $a \mid 12$ however there is no such condition on $c$ here, thus we can allow $a$ to take any of the 12 divisors of 12, and setting $c = 288/a^2-a$ does the job. So we have solutions $(-4,6,12)$ and $(d,d,288/d^2-d)$ for $d\mid 12$ . A total of $1+12 = \boxed{13}$ solutions.