smc

We observe that because $P(1)=P(3)=P(5)=P(7)=a$, if we define a new polynomial $R(x)$ such that $R(x)=P(x)-a$, $R(x)$ has roots when $P(x)=a$; namely, when $x=1,3,5,7$. Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x))=R(x)=P(x)-a$. Then, plugging in values of $2,4,6,8,$ we get $$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2)=-15Q(2)=-2a$$ $$P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4)=9Q(4)=-2a$$ $$P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6)=-15Q(6)=-2a$$ $$P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8)=105Q(8)=-2a$$ $-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $\text{lcm}(15,9,15,105)$. Solving, we receive $315$, so our answer is $\boxed{315}$. To complete the solution, we can let $a=315$, and then try to find $Q(x)$. We know from the above calculation that $Q(2)=42,Q(4)=-70,Q(6)=42$, and $Q(8)=-6$. Then we can let $Q(x)=T(x)(x-2)(x-6)+42$, getting $T(4)=28,T(8)=-4$. Let $T(x)=L(x)(x-8)-4$, then $L(4)=-8$. Therefore, it is possible to choose $T(x)=-8(x-8)-4=-8x+60$, so the goal is accomplished. As a reference, the polynomial we get is $$P(x)=(x-1)(x-3)(x-5)(x-7)((-8x+60)(x-2)(x-6)+42)+315$$ $$=-8x^7+252x^6-3248x^5+22050x^4-84392x^3+179928x^2-194592x+80325$$