Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Replies: 7   Last Post: Mar 15, 2011 2:36 AM

 Messages: [ Previous | Next ]
 qindars@gmail.com Posts: 45 Registered: 1/12/07
Posted: Mar 11, 2011 1:39 AM

Hi,

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
something called "excludents", which seems to be something like what I
am doing, but it's not totally clear to me.

Fran

Date Subject Author
3/11/11 qindars@gmail.com
3/11/11 Jose Carlos Santos
3/14/11 qindars@gmail.com
3/15/11 qindars@gmail.com
3/15/11 Tim Little
3/11/11 Rob Johnson
3/11/11 renu2010
3/13/11 Ilmari Karonen