jmc

Setting $x=10,$ we get $$(b+10)(c+10)=1.$$Either $b+10=c+10=1$ or $b+10=c+10=-1.$

If $b+10=c+10=1,$ then $b=c=-9,$ and $$(x-a)(x-10)+1=(x-9{)}^2.$$Since $(x-9{)}^2-1=(x-10)(x-8),$ $a=8.$

If $b+10=c+10=-1,$ then $b=c=11,$ and $$(x-a)(x-10)+1=(x-11{)}^2.$$Since $(x-11{)}^2-1=(x-12)(x-10),$ $a=12.$

Thus, the possible values of $a$ are $\boxed{8,12}.$