Question: The problem is to solve this:

$$a^2 + b^2 =c^4 \text{ }a,b,c\in \Bbb{N}\text{ }a,b,c<100$$

My idea: To see this problem I at first must see idea with Pythagorean triplets, and to problem is to find hypotheses of length square number. Is there any easier approach to this problem?

Answer: 
I think, the way with using Pythagorean triplets is the best.

Let $\gcd(a,b,c)=1$.

Thus, there are natural $m$ and $n$ with different parity such that $m>n$ and $\gcd(m,n)=1$ and
$a=2mn,$ $b=m^2-n^2$.

Thus, $c^2=m^2+m^2$ and by the same way there are naturals $p$ and $q$ with a different parity, such that $p>q$, $\gcd(p,q)=1$ and $m=2pq$ and $n=p^2-q^2$.

Thus, $a=4pq(p^2-q^2)$, $b=4p^2q^2-(p^2-q^2)^2=6p^2q^2-p^4-q^4$ and $c=p^2+q^2$.

Can you end it now?

For example, for $p=2$ and $q=1$ we obtain: $(a,b,c)=(24,7,5)$.
We need to find Pythagorean triples where the hypotenuse is already a square. We can certainly use $5\*(3,4,5)=(15,20,25)$ or $20\*(3,4,5)=(60,80,100)$ but we can also use the following formula to find one or more triples for a given hypotenuse, if they exist. For any $C,m$ that yields an integer, we have the $m,n$ we need for a triple with that hypotenuse.

For $C= m^2+n^2,\space n=\sqrt{C-m^2}$ where $\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor$. Let 

$$n=\sqrt{25-m^2}\text{ where }\biggl\lceil\sqrt{\frac{25}{2}}\space\space\biggr\rceil=4 \le m\le\bigl\lfloor\sqrt{25}\bigr\rfloor=5$$

We find $n(25,4)=3\rightarrow f(4,3)=(7,24,25)$ but $n(25,5)=0$; there is no triple for $m=5$

We could also find $(3,4,5)$ by using $C=5$, a factor of $25$

$$n=\sqrt{5-m^2}\text{ where }\biggl\lceil\sqrt{\frac{5}{2}}\space\space\biggr\rceil=2 \le m\le\bigl\lfloor\sqrt{5}\bigr\rfloor=2$$
We find $n(5,2)=1\rightarrow f(2,1)=(3,4,5)$

We find no integer values of $n$ for $C=49$ or $C=81$ or their factors

$$n=\sqrt{100-m^2}\text{ where }\biggl\lceil\sqrt{\frac{100}{2}}\space\space\biggr\rceil=8 \le m\le\bigl\lfloor\sqrt{100}\bigr\rfloor=10$$

We find $n(100,8)=6\rightarrow f(8,6)=(28,96,100)$ but $n(100,10)=0$; there is no triple for $m=10$