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: $(a,b,c)=(24,7,5)$.
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> $a^2 + b^2 =c^4 \quad a,b,c\in \Bbb{N}\quad a,b,c<100$
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and

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> Can you please comment how you get this solutions?
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The involved numbers are so small we can compute everything by hand and without brute-force search or such:

Due to $a, b < 100$ we have $c^4=a^2+b^2 < 2\cdot 100^2$ and thus 
$$c < \sqrt[4]{2}\cdot10<\sqrt{1.44}\cdot10=1.2\cdot10=12$$
where I used $\sqrt 2 < 1.44$. Hence $c$ is 11 at most.

Finding Pythagorean triples is equivalent to solving the norm equation in $\Bbb Z[i]$ with $i=\sqrt{-1}$, the [Gaussian integers](https://en.wikipedia.org/wiki/Gaussian_integer). As $1\leqslant c\leqslant 11$ we solve the norm equation for primes not 3 mod 4 and multiply the resulting prime ideals together to get the solutions. The only integer primes that decay over $\Bbb Z[i]$ are 2 and 5: $\def\p{\mathfrak p} \def\b{\bar{\p}}$

$2 = \p\_2 \b\_2$

$5 = \p\_5 \b\_5$

Real solutions like for for 0 and 1 are of no interest because you only want $a, b\in\Bbb N$. We then take the 4-th powers of these ideals (or their conjugate) to find the solutions for $c$. The only products that do not exceed 11 are: 2, 5 and 2·5=10 where 2 is of no interest because $\p\_2^{2n}=2^n$ is real and thus only gives unwanted solutions like $4^2 + 0^2 = 2^4$ (this is because $\p\_2=\b\_2$ modulo units where $\bar{\,\cdot\,}$ denotes complex conjugation).

We arrive at the following 4 solutions (modulo units and conjugation):
\begin{align}
s\_1 &= \p\_5^4 \\
s\_2 &= \p\_5^3\b\_5=5\p\_5^2 \\
s\_3 &= \p\_2^4\p\_5^4=2^2\p\_5^4 = 4s\_1\\
s\_4 &= \p\_2^4\p\_5^3\b\_5=2^25\p\_5^2 = 4s\_2
\end{align}
With $\p\_5 = 2+i$ we have $\p\_5^2=3+4i$ and $\p\_5^4=-7+24i$, thus:
\begin{array}{rclll}
s\_1 &=& -7+24i &\Rightarrow& (7,24,5)\\
s\_2 &=& 5(3+4i) &\Rightarrow&(15,20,5)\\
s\_3 &=& 4(-7+24i)&\Rightarrow&(28,96,10)\\
s\_4 &=& 20(3+4i)&\Rightarrow&(60,80,10)
\end{array}
These solutions are only up to ordering $a$ and $b$, i.e. $(24,7,5)$ etc. are also solutions, of course.