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Rah4927 , I know about what you wrote . Never ending was my first step . Disorder in sequence would have been the next . I just kind of couldn't do it and confused me well . Well , and thank you .
In p3 , let's have two odd numbers - 2n+1 and 2p+1 . Their sum = 2n+2p+2 = 2(n+p+1) . Suppose 2n+2p+2 is divisible by 2 , but not 4 . So n+p+1 is not divisible by 2 . The difference = 2n-2p = 2(n-p) . n+p+1 is not divisible by 2 . So n+p must be divisible by 2 . That can be possible in two ways . Ei...
- Sun Aug 30, 2015 9:26 am
- Forum: National Math Camp
- Topic: Exam 2, Online Number Theory Camp, 2015
- Replies: 24
- Views: 11920
Can p4 be solved like this ? (a^2 - 1)*(b^2-1) = a^2*b^2-a^2-b^2+1 . It is divisible by both a and b . And , a^2*b^2 is divisible by ab . So that leaves 1-a^2-b^2 . Let's multiply it with (-1) . Result = a^2+b^2+1 which is divisible by ab . And as for k being a positive integer , a and b are both gr...