prove that $X^2 \equiv 35 \pmod{100}$ has no solutions

Question

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.

Answer 1

$X^2=100k+35=4(25k)+35=4(25k+8)\color{red}{+3}$.

Answer 2

$100k+35$ 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.

Answer 3

If $ x^2 = 100*n + 35 $ which is multiple of 5, so x would be also multiple of 5. The left hand side would be multiple of 25, but the right hand side is $ 25*(4*n+1)+10 $ which is not multiple of 25. Hence the equation has no solutions.