prove that $X^2 \equiv 35 \pmod{100}$ has no solutions This problem is from  'A Survey of Modern Algebra' - Garret Birkoff, and Saunders Mac Lane in Section 1.9.
I'm an autodidact and there are no answers in the back so I need you guys to look at my proofs every once in a while to verify them. As you might be able to tell from my previous posts my mathematical maturity is not too high so I appreciate your patience. I give you permission to laugh at 
this post from 6 months ago but I'm getting better.
The truth of the congruence 
$X^2 \equiv 35 \pmod{100}$ can be inferred from looking at $100|(x^2-35)$ which has the equivalent of saying $100k = x^2 -35$ for some $k \in \Bbb{Z}$
by basic algebra $ x^2 =100k + 35 $ which I will now write as a function of k and attempt an induction.
Let
$P(k) = 100k + 35$
$P(0) = 35$ which is not a square.
$P(k+1) = 100(k + 1) + 35 = 100k + 100 +35 = (100k + 35) + 100$
by substitution from the induction hypothesis
$ P(0) + 100 = 35 + 100 = 135 $ 
which is also not a square. Q.E.D.

$100k+35$ is divisible by 5, so if it is a perfect square $a^2$, then $a$ is divisible by 5 too. But $a$ cannot be multiple of 10. On the other hand $(10b+5)^2=100b^2+100b+25$ which does not end in 35 either. 
