Some Challenging Problems:

1. Feb 1: If a,b,c are odd integers, prove that ax^2+bx+c = 0 does not have a rational number solution.
   Solution:
   uppose for the sake of contradiction that the statement is false. Then there is a rational number root p/q, where p,q are natural numbers and q > 0. We can further assume that p,q have no common factors.
   So a(p/q)^2 + b(p/q) + c = 0. Multiplying by q^2 throughout, we get ap^2 + bpq + cq^2=0.
   We need to consider four cases.
   Case 1 (p,q both odd): In this case all three terms are odd and three odd numbers cannot sum to 0.
   Case 2 (p odd, q even): In this case the first term is odd but the other two terms are even so that the sum must be odd and thus cannot be zero.
   Case 3 (p even, q odd): In this case the first two terms are even but the third is odd, so that the sum must be odd and thus cannot be zero.
   Case 4 (p,q both even): In this p,q must have a common factor of 2, which is not possible by our assumption.
   Thus in all cases we get a contradiction. So it must be the case that the given equation has no rational roots.
2. Is the following proposition true or false? Prove your answer. Let a,b,c be integers such that a^2+b^2=c^2. Then a is even or b is even.
3. Prove that for each real number x, (x + sqrt(3)) is irrational or (-x+sqrt(3)) is irrational.