Fermat's Last Theorem for $n=4$ states that
If $$b^4+c^4=a^4\tag{1}$$
Then, no positive integers $a, b, c$ exist simultaneously.
We can re-arrange (1) as
$$a^4-b^4=c^4\tag{2}$$
But, by direct expansion, we see that
$$a^4-b^4 = (a^2+b^2)^2-(b(a+b))^2-(b(a-b))^2\tag{3}$$
So, equating (2) and (3), we get
$$c^4 = (a^2+b^2)^2-(b(a+b))^2-(b(a-b))^2$$
$$(c^2)^2 = (a^2+b^2)^2-(b(a+b))^2-(b(a-b))^2$$
$$(c^2)^2+(b(a+b))^2+(b(a-b))^2 = (a^2+b^2)^2\tag{4}$$
We can see that (4) is a Pythagoras' quadruple.
Pythagoras' quadruple states that
If $$w=m^2+n^2-p^2-q^2$$
$$x=2(mq+np)$$
$$y=2(nq-mp)$$
$$z=m^2+n^2+p^2+q^2$$
Then
$$w^2+x^2+y^2=z^2\tag{5}$$
Where $gcd(w,x,y,z)=1$
Since $w,x,y,z$ are co-prime, then $w$ and $z$ are odd while $x$ and $y$ are even.
This shows that two of $b(a+b)$, $b(a-b)$ and $a^2+b^2$ must be even while the remaining one is odd.
Now, comparing (4) and (5), we can see that
$$(c^2)^2=m^2+n^2-p^2-q^2$$
Thus, $c$ is odd. And if $c$ is odd, then we see from (2) that $a$ and $b$ must be of opposite parity. But $b$ can't be odd because then, $b(a+b)$, $b(a-b)$ and $a^2+b^2$ will all be odd which does not conform with Pythagoras' quadruple.
If $a$ is odd and $b$ is even, we see that $b(a+b)$ is even, $b(a-b)$ is even and $a^2+b^2$ is odd which corresponds with Pythagoras' quadruple.
So, we are forced to conclude that $a$ is odd while $b$ is even.
Now, we can see that
$$(c^2)^2=m^2+n^2-p^2-q^2\tag{6}$$
$$a^2+b^2=m^2+n^2+p^2+q^2\tag{7}$$
$$b(a+b)=2(mq+np)\tag{8}$$
$$b(a-b)=2(nq-mp)\tag{9}$$
Adding (8) and (9), we have
$$b(a+b) + b(a-b)=2(mq+np)+ 2(nq-mp)$$
$$b(a+b) + b(a-b)=2q(n+m)+ 2p(n-m)\tag{10}$$
We can see that to make both expressions of (10) obviously equal, we only need to set $p=q=\frac{m}{2}$. By doing this, we have
$$b(a+b) + b(a-b)=2\frac{m}{2}(n+m)+ 2\frac{m}{2}(n-m)$$
$$b(a+b) + b(a-b)=m(n+m)+ m(n-m)\tag{11}$$
Comparing both sides of (11), we can see that $b=m$ and $a=n$. Specifically, $m$ is even while $n$ is odd.
Putting $p=q=\frac{m}{2}$ in (6) gives
$$(c^2)^2=m^2+n^2-(\frac{m}{2})^2-(\frac{m}{2})^2$$
$$(c^2)^2=m^2+n^2-\frac{m^2}{2}$$
$$(c^2)^2=\frac{m^2}{2}+n^2$$
Since $m$ is even, say $m=2m_1$, then
$$(c^2)^2=2{m_1}^2+n^2$$
$$c^4=2{m_1}^2+n^2\tag{12}$$
Also, putting $p=q=\frac{m}{2}$ in (7) gives
$$a^2+b^2=m^2+n^2+(\frac{m}{2})^2+(\frac{m}{2})^2$$
$$a^2+b^2=m^2+n^2+\frac{m^2}{2}$$
$$a^2+b^2=\frac{3m^2}{2}+n^2$$
Since $m$ is even, say $m=2m_1$, then
$$a^2+b^2=6{m_1}^2+n^2\tag{13}$$
The final jigsaw is now to prove that at least one of (12) and (13) does not have integers solution.
Putting $a=n$, $b=m=2m_1$ in (13), we have;
$$n^2+(2m_1)^2=6{m_1}^2+n^2$$
$$4m_1^2=6{m_1}^2$$
$$2m_1^2= 0$$
Thus we have that $m_1=m=p=q=b=0$ which means (1) becomes
$$0^4+c^4=a^4$$ which is trivial. This makes us conclude that no three natural numbers satisfy (1)\
LATEST DEVELOPMENT
Let's begin a new approach with (6) and (7).
From (7), we need to know that $a^2+b^2>m^2+n^2$ since $p$ and $q$ cannot be equal to zero simultaneously, because then, we will have a trivial pythagoras quadruple.
Now, suppose (6) and (7) could exist simultaneously, multiplying them together gives
$$c^2(a^2+b^2)=(m^2+n^2)^2-(p^2+q^2)^2$$
$$(ac)^2+(bc)^2=(m^2+n^2)^2-(p^2+q^2)^2$$
$$(ac)^2+(bc)^2+(p^2+q^2)^2=(m^2+n^2)^2\tag{8}$$
We can see that (8) is a new Pythagoras quadruple which is less than (5) since $a^2+b^2=m^2+n^2+p^2+q^2>m^2+n^2$.
Since (8) is a Pythagoras quadruple, we may say that
$$ac=m_1^2+n_1^2-p_1^2-q_1^2\tag{9}$$ since $a$ and $c$ are odd.
We should note from (6) that since $m^2+n^2-p^2-q^2$ is always a perfect square, $m_1^2+n_1^2-p_1^2-q_1^2$ is also a perfect square. Hence, (9) becomes
$$(a_1c_1)^2= m_1^2+n_1^2-p_1^2-q_1^2\tag{10}$$
and
$$m^2+n^2=m_1^2+n_1^2+p_1^2+q_1^2\tag{11}$$
$$p^2+q^2=2(m_1q_1+n_1p_1)\tag{12}$$
$$bc=2(n_1q_1-m_1p_1)\tag{13}$$
From (11), we see that $a^2+b^2>m^2+n^2>m_1^2+n_1^2$ since $p_1$ and $q_1$ cannot be equal to zero simultaneously, because then, we will have a trivial pythagoras quadruple.
Now, multiplying (10) and (11) again leads to another pythagoras quadruple. This means that $a^2+b^2$ will keep decreasing on every new Pythagoras quadruple derived (i.e $$a^2+b^2>m^2+n^2>m_1^2+n_1^2>...>m_r^2+n_r^2$$
However, $a^2+b^2$ can't decrease forever which means that (6) and (7) cannot exist simultaneously, hence a contradiction that they can.
Question: Is my new approach correct? If not, where is/are the flaw(s)?