https://math.stackexchange.com/questions/351491/integral-solutions-of-hyperboloid-x2y2-z2-1/709384#709384

Since squares leave remainders 0 and 1 modulo 4, we have $a=2\alpha, b = 2 \beta, c = 2k +1$, which leads to $ \alpha^2 + \beta ^2 = k(k+1)$. Now, $[a/2] + [c/2] = \alpha + k$. Suppose $2|\alpha, 2\not | k$. Then $2|\beta$ and so, $4|(k+1)$. Take a prime $p$ of the form $4s+3$ with odd degree of occurence in $k$. Since $k$ and $k+1$ are coprime, it has odd degree of occurence in $\alpha^ 2 + \beta^2$, contradiction. The other case $2 \not | \alpha, 2 | k$ is analogous: $\alpha^2 + \beta^2 \equiv 2 (mod 4)$, so $k \equiv 2 (mod 4)$, so $k+1 \equiv 3 (mod 4)$

Use the fact that a prime of the form 4k+3 doesn't divide b^2+1.

liaombro wrote: ..... The other case $2 \not | \alpha, 2 | k$ is analogous: $\alpha^2 + \beta^2 \equiv 2 (mod 4)$, so $k \equiv 2 (mod 4)$, so $k+1 \equiv 3 (mod 4)$ Let's use the hint given by Zawos. Since $k+1\equiv3 (mod 4)$, there is a prime $p=4m+3$, $m\ge0$, such that $p\big|k+1$, and hence form the equality $\alpha^2+\beta^2=k(k+1)$, we get $p\big|\alpha^2+\beta^2$. So, $\alpha^2\equiv-\beta^2 (mod p)$. Now, from the Fermat's little theorem: $$1\equiv\alpha^{4m+2}\equiv(\alpha^2)^{2m+1}\equiv(-\beta^2)^{2m+1}\equiv-\beta^{4m+2}\equiv-1 (mod p).$$Contradiction! Thanks to liaombro and Zawos.