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: $(a,b,c)=(24,7,5)$.
All Solutions are $$(a=7\land b=24\land c=5)\lor (a=15\land b=20\land c=5)\lor (a=20\land b=15\land
 c=5)\lor (a=24\land b=7\land c=5)\lor (a=28\land b=96\land c=10)\lor
 (a=60\land b=80\land c=10)\lor (a=80\land b=60\land c=10)\lor (a=96\land
 b=28\land c=10)$$