Show that the equation $2a^4+2a^2b^2+b^4=c^2$ hasn't an integer solution Question:

Prove that: if $a,b,c\in Z$,and $a\neq 0$ show that: the equation
  $$2a^4+2a^2b^2+b^4=c^2$$ has no solution.

My idea:
since
$$a^4+(a^2+b^2)^2=c^2$$
and I want use Pythagorean triple
and let  $$a^2=m^2-n^2. a^2+b^2=2mn, c=m^2+n^2,(m,n)=1$$
$$\Longrightarrow  m^2-n^2+b^2=2mn$$
$$\Longrightarrow (m-n)^2+b^2=2n^2$$
maybe my idea is wrong?
maybe the David H idea  is usefull?
since
$$2a^2(a^2+b^2)=(c-b^2)(c+b^2)$$
then $b$ and $c$ are either both even or both odd,and $c>b^2>b$
can usefull
But I can't continue.

Let $(a,b,c)$ be a solution of
$$ 2a^4 + 2a^2b^2 + b^4 = c^2.$$
By evaluating the different values of $a,b,c$ modulo $4$ (Edit: $8$), one can easily show that $a$ must be even. If $b$ would also be even, then $(\frac a2, \frac b2, \frac c4)$ is also a solution to the equation. Hence if $a\neq 0$ we may assume that $b$ is odd and thus $c$.
Now we use, what you already wrote about Pythagorean triples. Note that automatically
$$ a^2 = 2mn\quad\text{and}\quad a^2 + b^2 = m^2-n^2$$
because $a$ is even. We also deduce the equations
$$ 2mn +b^2 = m^2 - n^2 \iff (m+n)^2 + b^2 = 2m^2.$$
So $m$ odd because $b$ is odd. The equation is also equivialent to
$$\left(\frac{m+n-b}2\right)^2+\left(\frac{m+n+b}2\right)^2 = m^2.$$
Therefore there must be $x$ and $y$ with
$$ m= x^2+y^2,\quad n= 2y(x-y),\quad b= y^2-x^2+2xy.$$
Therefore
$$2mn = 4(x^2+y^2)y(x-y)$$
must be the square $a^2$. It should be easy to check that the three factors are pairwise coprime, thus
$$ x^2+y^2 = z^2,\quad y = w^2, \quad x-y = t^2.$$
Putting these equations together we get
$$ z^2 = (t^2 + w^2) + w^4 = 2w^4 + 2w^2t^2 + t^4.$$
Hence $(w,t,z)$ is also a solution of the above equation, but $|w|<|a|$. By infinite decent, there is no solution $(a,b,c)$ with $a\neq 0$.
(Note that one needs to pay more attention on coprimeness at a few steps which I hope is correct or can be worked around.)
