@Bl4nkB0x The proof given shows that when n=2m+1, then n^4-1= 8(2m^4+4m^3+3m^2+m). To prove divisibility by 16, you need to show that 2m^4+4m^3+3m^2+m is even. This is obvious for the first two terms (2m^4+4m^3), so you are left with the problem of proving that, for all integers m, 3m^2+m is even. Again this is obvious if m is even. But if m is odd, then so is 3m^2, and so 3m^2+m is a sum of two odd integers, hence is even.
@JoelFeinstein That's brilliant! I tried something similar: trying to show that for all integers m, 8(2m^4+4m^3+3m^2+m) is even. I tried to show separately for when m is odd and even, instead of separating the terms and seeing it's quite obvious to prove.
@elninovaquero This is the definition of "odd integer" we are working with. The even integers are integers of the form 2m for some other integer m, while the odd integers are integers of the form 2m+1 instead. (Equivalently, you can ask for remainder 1 when you divide by 2, but that is the same thing.)
How would you prove for divisible by 16?
Bl4nkB0x 2 months ago
@Bl4nkB0x The proof given shows that when n=2m+1, then n^4-1= 8(2m^4+4m^3+3m^2+m). To prove divisibility by 16, you need to show that 2m^4+4m^3+3m^2+m is even. This is obvious for the first two terms (2m^4+4m^3), so you are left with the problem of proving that, for all integers m, 3m^2+m is even. Again this is obvious if m is even. But if m is odd, then so is 3m^2, and so 3m^2+m is a sum of two odd integers, hence is even.
JoelFeinstein 2 months ago
@JoelFeinstein That's brilliant! I tried something similar: trying to show that for all integers m, 8(2m^4+4m^3+3m^2+m) is even. I tried to show separately for when m is odd and even, instead of separating the terms and seeing it's quite obvious to prove.
Thanks for your response.
Bl4nkB0x 1 month ago
How did you get the 2m+1??
elninovaquero 4 months ago
@elninovaquero This is the definition of "odd integer" we are working with. The even integers are integers of the form 2m for some other integer m, while the odd integers are integers of the form 2m+1 instead. (Equivalently, you can ask for remainder 1 when you divide by 2, but that is the same thing.)
JoelFeinstein 3 months ago