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NT III, 5. 25, 27, 28, 29

 
 
JinTina的头像
NT III, 5. 25, 27, 28, 29
JinTina - 2023年04月29日 Saturday 19:46
 

Hello!

Here are my thoughts for 5.29: I got that (2n-1) |  (n+1)^3, after crossing out terms with (2n-1) as a factor. Then, if n+1 and 2n-1 contain multiples of 3, 5 then they must be 14 (mod 15) or 2(mod 15). I don't really know what to do next. 

For 5.28, I tried polynomial division to get x^(mn-m)+x^(mn-2m)+..., but I don't really know what the remainder will be because I didn't find a pattern in the remainders after subtracting. 

For 5.27, I got x=a+4=b+1=c+1=d+1 and some other solutions similar to that one (by finding factors of 4) I don't think it is correct, but I can't seem to find any restrictions on the variables or anything like that. How can I solve the problem?


For 5.25, I got 7<n<21 by the triangle inequality, but can't seem to find any restrictions on m other than m<54, and m+21>33, so m>12 and m+33>21, so m>-12 which is intuitive. So then 12<m<54. But I don't know what restricts m when n is a given value. For example, if n=8, then what values of m would work??? 


Can you give me hints to the problems?


Thanks!


 
WangDr. Kevin的头像
Re: NT III, 5. 25, 27, 28, 29
WangDr. Kevin - 2023年05月1日 Monday 22:55
 

You are very close on 5.29.

For other questions, please look at my hints for the question under the following link:

https://classes.areteem.org/mod/forum/discuss.php?d=1293