An equation involving 2760.
Find the number of ordered triples of integers ( a , b , c ) that satisfy the equation a ( b − c ) 3 + b ( c − a ) 3 + c ( a − b ) 3 = 2 7 6 0 .
Note that the order matters: ( − 8 , − 5 , 1 5 ) and ( 1 5 , − 8 , − 5 ) are considered different solutions.
The answer is 66.
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2 solutions
I was so excited until I saw Mathematica... Looking for a better proof still :P I was stumped on the question so I skipped to solutions - anyone else have a solution?
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Follow the above, we get x y z ∣ 2 7 6 0 . WLOG, let x be the smallest in absolute value, then we can show that 2 7 6 0 ≥ ∣ x ∣ × ∣ x ∣ × ∣ 2 x ∣ so x ≤ 1 0 (finite cases). Next, you find suitable triples of ( x , y , z ) , and then check what values of ( a , b , c ) you get.
It is slightly tedious.
I did it the same way but without mathematica...it seemed tedios....wasted 2 hours....
python is way better method...
I used python to solve the problem
Continuing from Shamik Banerjee's solution
xy(x+y)(a+b+c)=2760, Plugging the values of x,y and z,we get- (a-b)(b-c)(c-a)(a+b+c)=2760
Here we can see that (a-b)+(b-c)+(c-a)=0 So we don't need to bother about a+b+c,we just have to make sure that 3 factors add to 0, a+b+c will take care of product of rest of the factors
2 7 6 0 = 2 3 × 3 × 5 × 2 3 So,the factors that qualify for the solution are( I have omitted the negative sign,You can put it to see how the sum is 0)
- ( 1 , 1 , 2 ) - 6 combinations{(-1,-1,2),(-1,2,-1),(2,-1,-1),(1,1,-2),(1,-2,1),(-2,1,1)}.
- ( 1 , 2 , 3 ) - 12 combinations.
- ( 1 , 4 , 5 ) )- 12 combinations.
- ( 1 , 5 , 6 ) - 12 combinations.
- ( 3 , 5 , 8 ) )- 12 combinations.
- ( 8 , 1 5 , 2 3 ) - 12 combinations.
Total 6 6 solutions
a (b-c)^3 + b (c-a)^3 + c (a-b)^3 = 2760 => (a-b) (b-c) (c-a) (a+b+c) = 2 2 2 3 5*23
Say, a-b = x; b-c = y; and c-a = z. x + y + z = 0 ==> z = - (x + y)
x y {- (x+y)} (a+b+c) = 2760 = 2 2 2 3 5 23 ==> x y (x+y) (a+b+c) = - 2 2 2 3 5 23
I used Mathematica to find out that there are 66 triples as it seemed unlikely that there is a short way to solve it