SOLUTION: Find all integers x for which x^3 = (x - 1)^3 + (x - 2)^3 + (x - 3)^3 + (x - 4)^3 + (x - 5)^3 + (x - 6)^3 + (x - 7)^3.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find all integers x for which x^3 = (x - 1)^3 + (x - 2)^3 + (x - 3)^3 + (x - 4)^3 + (x - 5)^3 + (x - 6)^3 + (x - 7)^3.      Log On


   

Question 1209692: Find all integers x for which x^3 = (x - 1)^3 + (x - 2)^3 + (x - 3)^3 + (x - 4)^3 + (x - 5)^3 + (x - 6)^3 + (x - 7)^3.
Answer by ikleyn(52915) About Me  (Show Source):
You can put this solution on YOUR website!
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Find all integers x for which x^3 = (x - 1)^3 + (x - 2)^3 + (x - 3)^3 + (x - 4)^3 + (x - 5)^3 + (x - 6)^3 + (x - 7)^3.
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Let's consider this equation

    x^3 = (x - 1)^3 + (x - 2)^3 + (x - 3)^3 + (x - 4)^3 + (x - 5)^3 + (x - 6)^3 + (x - 7)^3.    (1)


        I am going to prove that there are no such integer numbers, 
                   that satisfy this equation.


Let's consider the numbers modulo 3.


If x is divisible by 3, then the table for x-1, x-2, x-3, x-4, x-5, x-6 and x-7 mod 3 is this

         x-1      x-2     x-3       x-4       x-5      x-6      x-7  
 mod 3    -1        1       0        -1         1        0       -1


The table for (x-1)^3, (x-2)^3, (x-3)^3,  (x-4)^3, (x-5)^3, (x-6)^3 and (x-7)^3 mod 3 is this

        (x-1)^3  (x-2)^3  (x-3)^3   (x-4)^3  (x-5)^3  (x-6)^3  (x-7)^3
 mod 3    -1        1        0         -1       1        0        -1


The sum of remainders "mod 3"  is  -1 + 1 + 0 -1 + 1 + 0 -1 = -1,
while x^3 mod 3 is 0 in this case.


    +-------------------------------------------------------------+
    |    It means that the sum in the right side of equation (1)  |
    |            can not be equal to the left side:               | 
    |      they have different remainders when divided by 3.      |
    +-------------------------------------------------------------+



The same idea works for  x = 1 mod 3.  Indeed, then the tables are these 


         x-1      x-2     x-3       x-4       x-5      x-6      x-7  
 mod 3     0       -1       1         0        -1        1        0


The table for (x-1)^3, (x-2)^3, (x-3)^3,  (x-4)^3, (x-5)^3, (x-6)^3 and (x-7)^3 mod 3 is this

        (x-1)^3  (x-2)^3  (x-3)^3   (x-4)^3  (x-5)^3  (x-6)^3  (x-7)^3
 mod 3     0       -1        1         0       -1        1        0


The sum of remainders "mod 3"  is  0 +-1 + 1 + 0 - 1 + 1 + 0 = 0,
while x^3 mod 3 is 1 in this case.


    +-------------------------------------------------------------+
    |    It means that the sum in the right side of equation (1)  |
    |            can not be equal to the left side:               | 
    |      they have different remainders when divided by 3.      |
    +-------------------------------------------------------------+



The same idea works for  x = 2 mod 3.  Indeed, then the tables are these 


         x-1      x-2     x-3       x-4       x-5      x-6      x-7  
 mod 3     1        0      -1         1         0       -1        1


The table for (x-1)^3, (x-2)^3, (x-3)^3,  (x-4)^3, (x-5)^3, (x-6)^3 and (x-7)^3 mod 3 is this

        (x-1)^3  (x-2)^3  (x-3)^3   (x-4)^3  (x-5)^3  (x-6)^3  (x-7)^3
 mod 3     1        0       -1         1        0       -1        1


The sum of remainders "mod 3"  is  1 + 0 - 1 + 1 + 0 - 1 + 1 = 1,
while x^3 mod 3 is -1 in this case.


    +-------------------------------------------------------------+
    |    It means that the sum in the right side of equation (1)  |
    |            can not be equal to the left side:               | 
    |      they have different remainders when divided by 3.      |
    +-------------------------------------------------------------+


Thus, considering all possible case for (x mod 3), we proved that 
equation (1) can not be valid.


It this point, the problem is solved completely.


ANSWER.  The given equation has no solutions in integer numbers.

Solved.