26. If $a, b$ are positive integers, and satisfy $(a+b+3)^{2}=4\left(a^{2}+b^{2}\right)$, then the number of pairs $(a, b)$ that satisfy the equation is ( ).
(A). 0
(B). 2
(C). 4
(D). 6
Answer: $B$.
Solution: It is clear that $a+b+3$ is an even number, so $a, b$ are one odd and one even, and if $(a, b)$ satisfies the condition, then $(b, a)$ must also satisfy it.
Without loss of generality, assume $a$ is odd and $b$ is even,
Since
$$
(a+b+3)^{2}=4\left(a^{2}+b^{2}\right),
$$

Expanding both sides, we get
$$
a^{2}+b^{2}+9+2 a b+6 a+6 b=4 a^{2}+4 b^{2}
$$
$$
\begin{array}{c}
3 a^{2}+3 b^{2}-2 a b-6 a-6 b-9=0 \\
\left(a^{2}-2 a b+b^{2}\right)+\left(a^{2}-6 a+9\right)+\left(b^{2}-6 b+9\right)+a^{2}+b^{2}=9+9+9 \\
(a-b)^{2}+(a-3)^{2}+(b-3)^{2}+a^{2}+b^{2}=27,
\end{array}
$$

Clearly, $a \leqslant 5, b \leqslant 5$,
But if $a=5$, then
$$
a^{2}+(a-3)^{2}=25+4=29>27,
$$

Therefore,
$$
a \leqslant 4, \quad b \leqslant 4 .
$$

Since
$a$ is odd, $b$ is even,

Thus
$a$ is 1 or 3, $b$ is 2 or 4.
(1) If $a=1$, substituting into the original equation, we get $(b+4)^{2}=4\left(1+b^{2}\right)$,
When $b=2$, the left side equals 36, the right side equals 20, which does not hold;
When $b=4$, the left side equals 64, the right side equals 68, which does not hold.
(2) If $a=3$, substituting into the original equation, we get $(b+6)^{2}=4\left(9+b^{2}\right)$,
When $b=2$, the left side equals 64, the right side equals 52, which does not hold;
When $b=4$, the left side equals 100, the right side equals 100, which holds.

In summary, only $(3,4)(4,3)$ satisfy the condition, i.e., there are 2 pairs of $(a, b)$.

Therefore, the answer is B.