<s>[INST] Can $\sqrt{ab}$ and $\sqrt{a^2 + b^2}$ be both integers if $a$ and $b$ are natural numbers? Does there exist an $a \in \mathbb{N}$ and $b \in \mathbb{N}$ such that $\sqrt{ab} \in \mathbb{Z}$ and $\sqrt{a^2 + b^2} \in \mathbb{Z}$?
 [/INST] As shown in the answer of Will Jagy, the problem reduces to showing that the equation
$$
A^4+B^4=C^2, \tag{1}
$$ 
does not possess a positive integer solution.
However, the fact that $(1)$
does not possess positive integer solutions can NOT be derived Fermat's Last Theorem. On the contrary, the fact that $(1)$ does not possess positive integer solutions implies Fermat's Last Theorem for $n=4$!
Here is why there is no solution to the Diophantine equation $(1)$.
The proof is based on a construction of an infinite descent, i.e., assuming that $C$ is the least positive integer for which the exists a triplet of positive integers satisfying $(1)$, we shall construct another such triplet with a small $C$. 
Clearly, $(A^2,B^2,C)$ is a Pythagorean Triple (assuming without the loss of generality, $(A,B)=1$ and $B$ is even), so co-prime integers $a,b$ exist such that
$$
A^2=a^2-b^2,\quad
B^2=2ab\quad\text{and}\quad C=a^2+b^2.
$$
The first equation becomes $A^2+b^2=a^2$, so $(A,b,a)$ is another Pythagorean Triple, with $b$ even.  Thus co-prime integers $c,d$ exist such that
$$
A=c^2-d^2, \quad
b=2cd\quad\text{and}\quad\text{and}\quad
a=c^2+d^2.
$$
Substituting the expressions for $b$ to $B^2=2ab$, we have
$$
B^2=4cd(c^2+d^2).
$$
Now, since $c, d, c^2+d^2$ are clearly pairwise co-prime, they all are perfect squares. Hence,
$$
c=e^2, \quad d=f^2\quad \text{and}\quad c^2+d^2=g^2.
$$
Combining these expressions we obtain that
$$
e^4+f^4=g^2,
$$
with $g \le g^2 = a \le a^2 < C$.
Thus, assuming $C$ is minimally chosen, we found a smaller value for $C$.  
Therefore, no solution.
Note. The most elementary proof of Fermat's Last Theorem for $n=3$ is also based on infinite descent. See here.
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