Prove that if $a \equiv b\ (\text{mod}\ 26)$, then $a^2 \equiv 2ab - b^2\ (\text{mod}\ 4)$.
So what I have done is that took $a= b + 26k$, used the value of $a$ and plugged it into $a^2 \equiv 2ab - b^2\ (\text{mod}\ 4)$
so it became
$b^2 + 52kb + (26k)^2 = b^2 + 52kb$
I know that $52kb$ and $(26k)^2$ will become 0 if I do mod 4. I just dont know how to write this in more formal form. I am sure, I am missing something very trivial.
Trying to work both sides of a claimed equality at the same time is a pretty common way to get confused, in my experience.
So: knowing that we can say $a=b+26k$, let's examine the two halves of the intended equivalence in succession:
$\begin{align} a^2 &\equiv (b+26k)^2 \\ &\equiv b^2+52k+26^2k^2 \\ &\equiv b^2 \bmod 4 \\ \end{align}$
$\begin{align} 2ab-b^2 &\equiv 2b(b+26k) -b^2\\ &\equiv 2b^2+52kb-b^2 \\ &\equiv b^2 +52kb \\ &\equiv b^2 \bmod 4 \\ \end{align}$
Therefore $a^2 \equiv 2ab-b^2 \bmod 4$
There are other methods, but this uses your observed form of $a$.
Hint $\,\ 2\mid 26\mid a-b\,\Rightarrow\, 4\mid (a-b)^2 = a^2-2ab+b^2.\ $ Or, said without divisibility language:
$a-b = 2n\,\overset{\rm square}\Rightarrow\,a^2-2ab+b^2 = 4n^2\,$ so $\ a^2 = 2ab-b^2+\color{#c00}4n^2\equiv 2ab-b^2\!\pmod{\! \color{#c00}4}$