Question

lovely little problem:

given that n is an integer satisfying

(n-3)^3 + n^3 = (n+3)^3

show that n must be even, and that n^2 must be a factor of 54

deduce that there is no such integer n

Answer 1

given that m is an integer satisfying:

(m-6)^3 + m^3 = (m+6)^3

deduce that m is even and hence show that there is no such integer m

Answer 2

dain nevermind...

Answer 3

has anyone solved it?

Answer 4

does this have anything to do with the question or are these separate questions?
(m-6)^3 + m^3 = (m+6)^3

Answer 5

that is an extension to the question, if you have done the original question, try that one :)

Answer 6

ok

Answer 7

subtract (n-3)^3 from both side results in:
n^3 = 18n^2 + 54

the right side is clearly even so n^3 must be even... this implies n must be even.

Answer 8

yeah :) i did that bit by contradiction, assuming n was odd

Answer 9

how do you do that second part.. show n^2 is a factor of 54...

Answer 10

you're almost there with the last post,

each line is more of a hint, read as far as you like

n^3 - 18n^2 = 54

n^2(n - 18) = 54

n is an integer, so n -18 is an integer

hence n is a factor of 54

Answer 11

oh... DUH!! :)

Answer 12

i think i'm losing brain cells...

Answer 13

Refer to the attachment.

Answer 14

nice, although personally i prefer the method which doesnt require plotting a graph.

has anyone attempted the second part?

Answer 15

let x be an integer,
54 = 2 * 3 * 3 * 3

x^2 = +ve
(x-18) ==> x > 18 ==> x^2 > 324
we have assumed x is an integer, (x-18) is also a positive integer
x^2(x-18) > 324 <--- contradiction
therefore .... x is not an integer!!

Answer 16

Let's do the second question you posted.

If we expand everything out, combine like terms, and set equal to 0, we get \[m^3=36m^2+432=2(18m^2+216)\]Since the right side is even, so is the left side. So , \(2|m^3\) so \(2|m \implies m\) is even.

Like the other problem you posted, \(m^2(m-36)=432\) so it can be deduced that \(m^2|432\). Since \(432=2^4\cdot3^3\) the only possibilities for \(m^2\) are \(1, 4, 16, 9, 36, 144\). So \(m=\pm1, 2, 3, 4, 6, 12\).

Now let's go to the equation \(m^2(m-36)=432\). Notice that \(m^2 >0\), and \(432>0\). Hence, \(m-36\) must be positive. However, none of our positive solutions allow that. Hence, there are no integral solutions.

Answer 17

nice! i did it slightly differently, i'll post it

Answer 18

(n-3)^3 + n^3 = (n+3)^3

assume n is odd
we get:
even + odd = even
contradiction!
hence n must be even

expanding:

n^3 - 18n^2 = 54

n^2(n - 18) = 54

n is an integer, so n -18 is an integer

hence n^2 is a factor of 54

Answer 19

since n is an even integer, let n = 2k

n^2 = 4k^2

54 is not a multiple of 4 therefore we have a contradiction

in other words there is no integer n:

n^2(n-18) = 54

Answer 20

for part 2 i used the same technique to show m was even,
this time expanding gets:

m^3-36m^2=432

m^2(m-36) = 432

notice 432 = 8 x 54

let m = 2r since its even

4r^2(2r - 36) = 432

8r^2(r-18) = 432

r^2(r-18) = 54

and we are done because we showed no integers satisfy this in part 1