Inequality or Chinese Remainder Theorem?

The following method was used by Fermat, and is known as the infinite descent method $a^2 \equiv 0,1 \pmod{4}, \quad b^2 \equiv 0,1 \pmod{4}, \quad c^2 \equiv 0,1 \pmod{4}.$ This implies that $a^2b^2 \equiv 0 \pmod{4}.$ , It's not possible $a^2b^2 \equiv 1 \pmod{4}$ because if this were true $a^2b^2 \neq a^2 + b^2 + c^2 \text{ mod 4 }$ .

Then there exists 5 possibilities. I'm going only to treat the following possibility: $a = 4k, b = 4l, c = 4m \Rightarrow 16(l^2 + k^2 + m^2) = 256l^2k^2 \Rightarrow l^2 + k^2 + m^2 = 16k^2l^2$ .

Continuing this method, descending we'll get the equation WLOG $1^2 + 1^2 + c^2 = 16^n, \space n \ge 1$ , but this equation doesn't have any integer $c$ fullfiling it. So the equation above doesn't have any solution with $a,b,c$ positive integers...

Note.- There are 4 cases more for analyzing. I hope to finish this proof... Meanwhile the other 4 cases are left as an exercise to the reader

Here's are a solution:

After some algebraic manipulation we get the following:

$a^2 + b^2 + c^2 = a^2b^2 \implies \left(a-1\right)\left(a+1\right)\left(b-1\right)\left(b+1\right) = c^2 + 1$

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Now if we have an odd prime $p$ which divides the LHS, then $-1$ must be a quadratic residue $\pmod{p}$ .

Since $\left(\dfrac{-1}{p}\right)_L \equiv \left(-1\right)^{\frac{p-1}{2}} \pmod{p}$ . (Using Legendre symbol and Euler's criterion )

$p \equiv 1 \pmod{4}$ must be true if $p$ is an odd prime.

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However, if either $a$ or $b$ is even, then the LHS is divisible by a positive integer which leaves a remainder of $3$ when divided by $4$ and thus is divisible by a prime which leaves a remainder of $3$ when divided by $4$ , which implies that $a,b\text{ and } c$ must be odd.

Now, $a = 2x + 1, b = 2y + 1, c = 2z + 1$ for non-negative integers $x,y,z$ .

After substitution and some algebraic manipulation we get, $8x\left(x+1\right)y\left(y+1\right) = z\left(z+1\right) + 1$ , where LHS is clearly even and RHS is clearly odd.

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Hence, there are no positive integers $a,b,c$ which solve this diophantine equation. $\square$