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find all integers n for which n^3=(n-1)^3+(n-2)^3+(n-3)^3

 

I don't think that the cubic formula will help, but I'm not sure

 Jun 6, 2020
edited by Guest  Jun 6, 2020
 #1
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n = 6

 Jun 6, 2020
 #2
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Guest's answer is correct.

 

If you wish to do this problem algetraicly:

(n - 1)3  =  n3 - 3n2 + 3n - 1

(n - 2)3  =  n3 - 6n2​ + 12n - 8

(n - 3)3  =  n3 - 9n2​ + 27n - 27

Therefore:

(n - 1)3 + (n - 2)3 + (n - 3)3  =  [ n3 - 3n2 + 3n - 1 ] + [ n3 - 6n2​ + 12n - 8 ] + [ n3 - 9n2​ + 27n - 27 ]

                                            =  3n3 - 18n2​ + 42n - 36

 

Since n3  =  (n - 1)3 + (n - 2)3 + (n - 3)3,

          n3  =  3n3 - 18n2​ + 42n - 36

            0  =  2n3 - 18n2​ + 42n - 36

            0  =  n3 - 9n2​ + 21n - 18

 

Try various factors of -18 to find an integer divisor of  n3 - 9n2​ + 21n - 18.

Sooner, or later, you will get to  n = 6.

Since  n = 6  is a divisor,  n3 - 9n2​ + 21n - 18  =  (n - 6)(n2 - 3n + 3)

 

Since  n2 - 3n + 3  cannot be factored using only integers, the only answer is:  n = 6.

 Jun 6, 2020

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