<s>[INST] 3 Questions on number theory. We have to everything using number congruences, and I am just a beginner, I know a few theorems, and we have to solve these using basics.  
1) If $n=a^4$ where $a \in \mathbb Z$ then prove that $n \equiv 0,1,5$ or $6(mod \ 10)$.
My work: $a^4 \equiv x(mod \ 10)$
We have to find $x$.
But how to simplify the above congruence ?  
2) Prove that $4n^2+4 \equiv 0(\mod \ 19)$ for any $n$.
But I think this question is wrong because it doesn't work for $1$ or $2$. I think it may be a misprint. It may be $\not\equiv$ in place of $\equiv$  
3) Solve for $n$, $5n \equiv 3(mod \ 8)$.
Sol: $n \equiv -1(mod \ 8)$
So $n=8k-1$
Solution set becomes $\{...,-17,-9,-1,7,15,...\}$
 [/INST] Question 1:  Because we can write any number $a \in \mathbb{Z}$ as $a=10\cdot b + c$ with $b,c \in \mathbb{N}$ we see that $a^4 = (10 \cdot b + c)^4 \mod 10$. It is easy to see that this forms a polynomial with all terms except for the $c^4$ part is a multiple of $10$. Since we are working in $\mod 10$ we see that $a^4=(10 \cdot b + c)^4 = c^4 \mod 10$. This means we only have to look at the last digit of any number in $\mathbb{Z}$ and we are left with 10 cases: 


*

*Numbers ending on $0: 0^4=0$ so last digit is $0$. 

*Numbers ending on $1: 1^4=1$ so last digit is $1$.

*Numbers ending on $2: 2^4=16$ so last digit is $6$.

*Numbers ending on $3: 3^4 =81$ so last digit is $1$.

*Numbers ending on $4: 4^4 =256$ so last digit is $6$.

*Numbers ending on $5: 5^4 =625$ so last digit is $5$.

*Numbers ending on $6: 6^4 =1296$ so last digit is $6$.

*Numbers ending on $7: 7^4 =2401$ so last digit is $1$.

*Numbers ending on $8: 8^4 =4096$ so last digit is $6$.

*Numbers ending on $9: 9^4 =6561$ so last digit is $1$.


Thus we have the numbers $0,1,5,6$.
Question 2: 
$4n^2+4 = 0 \mod 19$ is not true, take $n=1$ and the result is trivial.
If we instead look at $4n^2+4 \neq \mod 19$ we see that we get a repeating pattern every 19 numbers. This can be seen by looking at \begin{align*}
4(n+19)^2+4 \mod 19 &= 4(n^2 + 38n + 19^2) +4 \mod 19 \\
&= 4 n^2 +4 \mod 19 + 38n \mod 19 + 19^2 \mod 19\end{align*} where the last 2 terms are obviously $0 \mod 19$.
You now have to compute the first 19 cases by hand (or excel/program it like I did), which shows indeed that $4n^2+4 \neq \mod 19$
Question 3: You have done this correctly.
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