Exercise 6 on page 26 of Birkhoff and MacLane's "A Survey of Modern Algebra" (2nd edtn), asks to show that x^2 cannot be congruent to 35 mod 100.
I think I can show this to be true, but I wonder if my way is the best one. I am wondering if anyone has a nicer/cleaner way to solve this problem.
I start off by supposing there is an integer x such that x^2 is congruent to 35 mod 100, which would imply that there exists an integer n such that x^2 = 100*n + 35. Since 100*n + 35 = 5*(20*n+7), this implies that 5|x^2 and since 5 is prime this implies further that 5|x, hence x can be written as x=5y. Substituting this form for x back into the previous eqn implies 5*y^2 = 20*n + 7 implies 5*(y^2-4*n) = 7 implies 5|7, which we know is false.
What bothers me I guess is that my approach seems very to be of a somewhat "ad hoc" nature. What if the problem had 63 or 64 or 32 instead of 35. Would I have to go through the same type of argument and would this type of argument even always work? I looked up solving quadratic congruences on the net and Wolfram's site talks about something called "excludents", which seems to be something like what I am doing, but it's not totally clear to me.
Thanks for any insights/help you can provide, Fran