$\because a^{2}+2a+40$是完全平方数，
$\therefore$设$a^{2}+2a+40=k^{2}$，其中$k$为正整数，
$\therefore \left(a+1\right)^{2}+39=k^{2}$，
$\therefore \left(a+1\right)^{2}-k^{2}=-39$，
即$\left(a+1+k\right)\left(a+1-k\right)=-39$，
$\because a$，$k$均为正整数，
$\therefore a+1+k$和$a+1-k$均为正整数，且$a^{2}+2a+40 \gt 40$，
$\therefore k^{2} \gt 40$，
$\therefore k \gt 6$且为整数，
$\therefore a+1+k \gt 7$，
$\because \left(a+1+k\right)\left(a+1-k\right)=-39=-1\times 39=-3\times 13$，
$\therefore$①$\left\{\begin{array}{l}{a+1+k=39}\\{a+1-k=-1}\end{array}\right.$或②$\left\{\begin{array}{l}{a+1+k=13}\\{a+1-k=-3}\end{array}\right.$，
由①解得：$\left\{\begin{array}{l}{a=18}\\{k=20}\end{array}\right.$，由②解得：$\left\{\begin{array}{l}{a=4}\\{k=8}\end{array}\right.$，
$\therefore a$的值是$18$或$4$.

$\because a^{2}+2a+40$是完全平方数，
$\therefore$设$a^{2}+2a+40=k^{2}$，其中$k$为正整数，
$\therefore \left(a+1\right)^{2}+39=k^{2}$，
$\therefore \left(a+1\right)^{2}-k^{2}=-39$，
即$\left(a+1+k\right)\left(a+1-k\right)=-39$，
$\because a$，$k$均为正整数，
$\therefore a+1+k$和$a+1-k$均为正整数，且$a^{2}+2a+40 \gt 40$，
$\therefore k^{2} \gt 40$，
$\therefore k \gt 6$且为整数，
$\therefore a+1+k \gt 7$，
$\because \left(a+1+k\right)\left(a+1-k\right)=-39=-1\times 39=-3\times 13$，
$\therefore$①$\left\{\begin{array}{l}{a+1+k=39}\\{a+1-k=-1}\end{array}\right.$或②$\left\{\begin{array}{l}{a+1+k=13}\\{a+1-k=-3}\end{array}\right.$，
由①解得：$\left\{\begin{array}{l}{a=18}\\{k=20}\end{array}\right.$，由②解得：$\left\{\begin{array}{l}{a=4}\\{k=8}\end{array}\right.$，
$\therefore a$的值是$18$或$4$.