1. Given positive integers $a$, $b$, $c$, $d$ satisfy $a^{2}=c(d+29)$, $b^{2}=c(d-29)$. Then the value of $d$ is $\qquad$.
$=, 1.421$
Let $(a, b)=d^{\prime}, a=d^{\prime} a_{1}, b=d^{\prime} b_{1},\left(a_{1}, b_{1}\right)=1$.
From the given conditions, dividing the two equations yields
$$
\begin{array}{l}
\frac{a^{2}}{b^{2}}=\frac{d+29}{d-29} \Rightarrow \frac{a_{1}^{2}}{b_{1}^{2}}=\frac{d+29}{d-29} \\
\Rightarrow\left\{\begin{array}{l}
d+29=a_{1}^{2} t, \\
d-29=b_{1}^{2} t
\end{array}\left(t \in \mathbf{Z}_{+}\right)\right. \\
\Rightarrow 2 \times 29=\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right) t .
\end{array}
$$

Since $a_{1}+b_{1}$ and $a_{1}-b_{1}$ have the same parity,
$$
t=2, a_{1}+b_{1}=29, a_{1}-b_{1}=1 \text {. }
$$

Thus, $a_{1}=15, b_{1}=14, d=421$.