Question 1209692
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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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<pre>
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.


<U>ANSWER</U>.  The given equation has no solutions in integer numbers.
</pre>

Solved.