not by induction. I did not post this in the original thread, because I did not want to hijack it.
I don't think the original problem as asked by the OP there should ask to do it by induction,
because I find the solution to be unwieldy.
(I thank Harry_the_cat for his multiple posts of explaining and working out many steps there.)
\(\displaystyle n^5 - n \ = \ \)
\(\displaystyle n(n^4 - 1) \ = \ \)
\(\displaystyle n(n^2 - 1)(n^2 + 1) \ = \ \)
\(\displaystyle n(n - 1)(n + 1)(n^2 + 1) \ = \ \)
\(\displaystyle (n - 1)(n)(n + 1)(n^2 + 1) \)
The expression is divisible by 3, because it is the product of three consecutive integers.
If the odd number ends in a 3 or 7, then its square will end in a 9. When added to 1 it
will be divisible by 5. So, the \(\displaystyle \ (n^2 + 1) \ \) factor will be divisible by 5.
If the odd number ends in a 1, then (n - 1) will be divisible by 5, if the odd number ends
in a 5, then n will be divisible by 5, or if the odd number ends in a 9, then (n + 1) will
be divisible by 5. So, in all cases of odd numbers, the expression is divisible by 5.
Let \(\displaystyle \ n = 2m + 1, \ \ \)where m is an integer. Substitute that into the expression:
\(\displaystyle [(2m + 1) \ - \ 1](2m + 1)[(2m + 1) \ + \ 1][(2m + 1)^2 \ + \ 1] \ = \ \)
\(\displaystyle (2m)(2m + 1)(2m + 2)(4m^2 + 4m + 2) \ = \ \)
\(\displaystyle (2)(2)(2)(m)(2m + 1)(m + 1)(2m^2 + 2m + 1) \ = \ \)
\(\displaystyle 8(m)(m + 1)(2m + 1)(2m^2 + 2m + 1) \)
The last expression is divisible by 8. Also, the last expression is divisible by an additional
2, because the product (m)(m + 1) is divisible by 2 as it is the product of two consecutive
integers. So, the expression is divisible by 16.
Taken together, the expression is divisible by 3, 5, and 16.
3*5*16 = 240
Therefore, \(\displaystyle \ n^5 - n \ \) is divisible by 240 when n is a positive odd integer.
I don't think the original problem as asked by the OP there should ask to do it by induction,
because I find the solution to be unwieldy.
(I thank Harry_the_cat for his multiple posts of explaining and working out many steps there.)
\(\displaystyle n^5 - n \ = \ \)
\(\displaystyle n(n^4 - 1) \ = \ \)
\(\displaystyle n(n^2 - 1)(n^2 + 1) \ = \ \)
\(\displaystyle n(n - 1)(n + 1)(n^2 + 1) \ = \ \)
\(\displaystyle (n - 1)(n)(n + 1)(n^2 + 1) \)
The expression is divisible by 3, because it is the product of three consecutive integers.
If the odd number ends in a 3 or 7, then its square will end in a 9. When added to 1 it
will be divisible by 5. So, the \(\displaystyle \ (n^2 + 1) \ \) factor will be divisible by 5.
If the odd number ends in a 1, then (n - 1) will be divisible by 5, if the odd number ends
in a 5, then n will be divisible by 5, or if the odd number ends in a 9, then (n + 1) will
be divisible by 5. So, in all cases of odd numbers, the expression is divisible by 5.
Let \(\displaystyle \ n = 2m + 1, \ \ \)where m is an integer. Substitute that into the expression:
\(\displaystyle [(2m + 1) \ - \ 1](2m + 1)[(2m + 1) \ + \ 1][(2m + 1)^2 \ + \ 1] \ = \ \)
\(\displaystyle (2m)(2m + 1)(2m + 2)(4m^2 + 4m + 2) \ = \ \)
\(\displaystyle (2)(2)(2)(m)(2m + 1)(m + 1)(2m^2 + 2m + 1) \ = \ \)
\(\displaystyle 8(m)(m + 1)(2m + 1)(2m^2 + 2m + 1) \)
The last expression is divisible by 8. Also, the last expression is divisible by an additional
2, because the product (m)(m + 1) is divisible by 2 as it is the product of two consecutive
integers. So, the expression is divisible by 16.
Taken together, the expression is divisible by 3, 5, and 16.
3*5*16 = 240
Therefore, \(\displaystyle \ n^5 - n \ \) is divisible by 240 when n is a positive odd integer.
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