need help on the following two questions
1.
2.
Thanks
We must have since 1 is a perfect square. Thus n can be written as a product of distinct primes: .
Now if all the ’s were even, n would be a perfect square. Since it isn’t, at least one of the ’s must be odd. Hence, we can write where m is odd and p does not divide k.
Suppose is rational: , where , and .
Then
as p is prime as .
Hence the factor of p occurs an odd number of times in . This is a contradiction because is a perfect square and so each prime factor of has to occur an even number of times.
Thus must be irrational.
It suffices to prove that has a solution with .
Let .
Let .
If then and . If then it means so . The negative sign is impossible so . Thus distinct elements in that set are not congruent to eachother. Thus, . Using a similar argument we can show . This means altogether and has elements. Thus, by pigeonhole principle there is and such that because the set of remainders mod is size . Thus, .