What's so special about 2760

Note that permuting any two of the numbers $a,b,c$ changes the sign of the polynomial $a(b-c)^3+b(c-a)^3+c(a-b)^3.$ In particular, if $a=b$ , then this expression is zero. This suggests that this polynomial is divisible by $a-b$ and, similarly, $a-c$ and $b-c.$ Indeed, the following beautiful identity is true: $a(b-c)^3+b(c-a)^3+c(a-b)^3=(a-b)(b-c)(c-a)(a+b+c)$ Note that permuting the three numbers $a,b,c,$ cyclically and/or simultaneously changing their signs does not change the value of this expression. Using these transformations, we can change any triple $(a,b,c)$ we are looking for into a triple for which $a+b+c>0$ and $a$ is the smallest of the numbers $a,b,c.$ Because $a-b<0,$ $c-a>0$ and $2760>0,$ this implies that $b-c<0.$ Therefore $a < b < c$ . Note that $a$ and/or $b$ may be negative, as long as $a+b+c>0.$ Each such triple will produce $6$ solutions by cyclic permutations and sign changes: $(a,b,c,),(c,a,b),(b,c,a),(-a,-b,-c),(-c,-a,-b),(-b,-c,-a)$ Clearly, all solutions are obtained in this manner, and they are distinct, because no two of the numbers $a,b,c$ are equal.

Suppose $b-a=x,\ c-b=y, c-a=z$ . Then all these numbers are positive, $z=x+y$ , and $xyz|2760=2^3\cdot 3\cdot 5 \cdot 23.$ The number $2760$ has $32$ divisors, listed below in the increasing order: $1,2,3,4,5,6,8,10,12,15,20,23,24,30,40,46,60,69,92,$ $115,120,138,184,230,276,345,460,552,690,920,1380,2760$

For the purpose of finding these $x,y,z,$ we can freely switch $x$ and $y$ . So suppose $x\leq y.$ Then, because $2760=xyz \geq x\cdot x\cdot (2x),$ we get $x\leq \sqrt[3]{1380},$ so $x\leq 10.$ We consider several cases, depending on what $x$ can be.

Case 1. $x=1$ . Then $y$ and $z$ must be consecutive divisors of $2760$ . We have six such triples $(x,y,z)$ with $x\leq y$ : $(1,1,2),(1,2,3),(1,3,4),(1,4,5),(1,5,6),(1,23,24).$

Case 2. $x=2$ . Then $y$ cannot be even, otherwise $z$ is also even and either $y$ or $z$ is divisible by $4$ . This would makes $xyz$ divisible by $2^4,$ which is impossible since $xyz$ divides $2760.$ This greatly reduces the number of $y$ to consider, and we get one triple: $(2,3,5).$

Case 3. $x=3.$ Then $y$ is not divisible by $3$ and $yz|920,$ so $y\leq 30$ . We get the following triples: $(3,5,8),(3,20,23).$

Case 4. $x=4.$ Then $y$ must be odd, and $yz|690,$ so $y\leq 23.$ There are no solutions in this case.

Case 5. $x=5.$ Then $y$ is not divisible by $5$ , and $yz| 552,$ so $y\leq 12.$ There are no solutions in this case.

Case 6. $x=6.$ Like in Case 2, $y$ is odd and so is $z$ . So $yz|115,$ there are no solutions in this case.

Case 7. $x=8.$ Then $y$ is odd and $yz|345,$ so $y\leq 15.$ We get one triple: $(8,15,23).$

Case 8. $x=10.$ Then $y$ is odd and $y\leq \sqrt{276},$ there are no solutions in this case.

Case 9. $x=12.$ Then $y$ is odd and $y\leq \sqrt{230},$ there are no solutions in this case.

All together, we get the following ten choices for $(x,y,z),$ up to switching $x$ and $y$ : $(1,1,2),(1,2,3),(1,3,4),(1,4,5),(1,5,6),(1,23,24),$ $(2,3,5),(3,5,8),(3,20,23),(8,15,23)$

We consider these cases separately. Most of them are similar, with the notable exceptions of $(1,1,2)$ and $(1,4,5).$

If $(x,y,z)=(1,1,2),$ then for some integer $k$ we must have $a=k-1,$ $b=k,$ $c=k+1$ . The sum $a+b+c$ equals $2760/(xyz)=1380.$ This implies $k=460,$ so we get a triple $(a,b,c)=(459,460,461).$

If $(x,y,z)=(1,2,3)$ or $(x,y,z)=(2,1,3)$ then for some $k$ $(a,b,c)=(k-1,k,k+2)$ or $(a,b,c)=(k-2,k,k+1).$ In both cases the sum must equal $2760/6=460.$ Because $460 \equiv 1 \pmod 3,$ in order for the number $k$ to be integer we must have $(a,b,c)=(k-1,k,k+2).$ This gives $k=153$ and a triple $(152,153,155).$

Similarly to the last case, for each of the remaining triples except $(1,4,5),$ exactly one of the two possibilities works modulo $3$ , and we get the following triples $(a,b,c):$

If $(x,y,z)=(1,3,4),$ then $(a,b,c)=(75,76,79).$

If $(x,y,z)=(5,1,6),$ then $(a,b,c)=(27,32,33).$

If $(x,y,z)=(23,1,24),$ then $(a,b,c)=(27,32,33).$

If $(x,y,z)=(3,2,5),$ then $(a,b,c)=(28,31,33).$

If $(x,y,z)=(3,5,8),$ then $(a,b,c)=(4,7,12).$

If $(x,y,z)=(3,20,23),$ then $(a,b,c)=(-8,-5,15).$

If $(x,y,z)=(8,15,23),$ then $(a,b,c)=(-10,-2,13).$

The case $(1,4,5)$ is slightly different, because both $(1,4,5)$ and $(4,1,5)$ are possible, producing triples $(44,45,49)$ and $(43,47,48).$ (What made this case different is that neither of the numbers $x,y,z$ is divisible by $3$ ).

Putting it all together, we have got $11$ triples, and each of them gives $6$ solutions by cyclic permutation and sign changes. So the answer is $66.$