jmc

$$\begin{array}{rl}[(a^2+2^a)+a\cdot 2^{(a+1)/2}][(a^2+2^a)-a\cdot 2^{(a+1)/2}]& =(a^2+2^a{)}^2-a^2\cdot 2^{a+1}\\ & =a^4+2\cdot a^2{2}^a+2^{2a}-a^2\cdot 2^{a+1}\\ & =a^4+2^{2a}\end{array}$$ (If you recall the reverse of Sophie Germain Identity with $a=a,b=2^{(a-1)/2}$, then you could have directly found the answer). By Fermat's Little Theorem, we have that $a^4\equiv 1(\mathrm{mod}5)$ if $a\nmid 5$ and $a^4\equiv 0(\mathrm{mod}5)$ if $a|5$. Also, we note that by examining a couple of terms, $2^{2a}\equiv 4(\mathrm{mod}5)$ if $a\nmid 2$ and $2^{2a}\equiv 1(\mathrm{mod}5)$ if $a|2$. Therefore,$$a^4+2^{2a}\equiv \{0,1\}+\{1,4\}\equiv \{0,1,2,4\}(\mathrm{mod}5)$$With divisibility by $5$ achievable only if $a\nmid 2,5$. There are $\frac{251-1}{2}+1=126$ odd numbers in the range given, and $\frac{245-5}{10}+1=25$ of those are divisible by $5$, so the answer is $126-25=\boxed{101}$.