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Cubic Equation (Posted on 2023-12-17) Difficulty: 3 of 5
Find all triplets (a, b, c) of integers a, b and c such that
       987654321a3 + 123456789b3 + c3 = (a + b + c)3
There are (at least) 3 solutions with a and b non-zero.

  Submitted by K Sengupta    
Rating: 5.0000 (1 votes)
Solution: (Hide)
The secret to solving this equation is realizing that 987,654,321 – 8(123,456,789) = 9.
 Let 987,654,321= A and 123,456,789 = B.
Then, nA – 8nB= 9n, where x = n and y = -8n. 
Let x+ y + z = m. 
Then, 9n + z = m^3 and n - 8n + z = z – 7n = m. Eliminating z gives 16n = m^3 – m = m(m – 1)(m + 
1), which is solvable in integers if 16 
divides m(m – 1)(m + 1). 
Then, n = 7+8k, where k is a positive integer.
Therefore, 16* n= (8k+7)(8k+6)(8k+8)
=> n = (8k+7)(4k+3)(k+1)
Accordingly, there are an infinite number of solutions,  
with:
(x,y,z) = (n, -8n, m+7n), where:
n=(8k+7)(4k+3)(k+1), m=7+8k 
k=1 gives: m = 15, n = 15*7*2 =210
x= 210, y = -8*210=-1680 z=15 -220+1680 = 1485
k= 2 gives: m = 23, n= 23*11*3  = 759
x= n = 759, y =-8*759 = -6072, z = 23-759+6072= 5336
k=3 gives: n= 31*15*4 =1860
x= n= 1860, y =-8*1860 = -14880, z = 31-1860+14880 = 13051

Consequently, summarizing the foregoing results, we have:
--> The given equation admits of an infinite number of solutions given by:
(x,y,z) = (n, -8n, m+7n), where:
n=(8k+7)(4k+3)(k+1), m=7+8k 
--> Solutions having the  three smallest values of x+y+z is given by: 
(x, y, z) = (210, -1680, 1485), (759, -6072, 5336), (1860, -14880, 13051)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
No SubjectK Sengupta2023-12-26 07:01:32
No SubjectLana Slater2023-12-26 04:56:33
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