1. Consider the quadratic equation

$(1+m^2)x^2-2(1+3m)x+(1+8m)=0.$

For a quadratic equation $ax^2+bx+c=0$ to have no real root, we need:

• $a \neq 0$, and
• discriminant $D<0$.

Here,

$a=1+m^2,\quad b=-2(1+3m),\quad c=1+8m.$

Since

$1+m^2>0$

for every integral $m$, the equation is always quadratic.

2. Compute the discriminant:

$D=b^2-4ac.$

So,

$D=[-2(1+3m)]^2-4(1+m^2)(1+8m).$

That is,

$D=4(1+3m)^2-4(1+m^2)(1+8m).$

Factor out $4$:

$D=4\left[(1+3m)^2-(1+m^2)(1+8m)\right].$

3. Expand the terms inside the bracket:

$(1+3m)^2=1+6m+9m^2.$

Also,

$(1+m^2)(1+8m)=1+8m+m^2+8m^3.$

Therefore,

$D=4\left[(1+6m+9m^2)-(1+8m+m^2+8m^3)\right].$

Simplify:

$D=4(-2m+8m^2-8m^3).$

Factor further:

$D=4[-2m(1-4m+4m^2)]$

and since

$1-4m+4m^2=(2m-1)^2,$

we get

$D=-8m(2m-1)^2.$

4. For no real roots, require

$D<0.$

Thus,

$-8m(2m-1)^2<0.$

Now, $(2m-1)^2\ge 0$ for all integers $m$, and it is zero only when $m=\tfrac12$, which is not an integer. Hence for integral $m$,

$(2m-1)^2>0.$

So the sign of $D$ depends only on $-m$.

Therefore,

$D<0 \iff m>0.$

5. Count integral values of $m$:

All positive integers satisfy the condition:

$m=1,2,3,\dots$

Hence there are infinitely many integral values of $m$.

6. Check options:

• A: $2$ ❌
• B: infinitely many ✅
• C: $1$ ❌
• D: $3$ ❌

Therefore, the correct option is B.