$q(x)=p(x)+x-3$ has integer coefficients and $q(\sqrt{2}+1)=0$ so $q(x)=(x-1-\sqrt{2})(x-1+\sqrt{2})r(x)=(x^2-2x-1)r(x)$ and $p(x)=(x^2-2x-1)r(x)-x+3$ $p(\sqrt{2}+2)= (1+2\sqrt{2})r(x)-\sqrt{2}+1$ $r(2+\sqrt{2})=c+d\sqrt{2}$ for some integer $c,d$ $a=(1+2\sqrt{2})(c+d\sqrt{2}) -\sqrt{2}+1= (2c+d-1)\sqrt{2}+c+4d+1$ so $2c+d-1=0$ or $d=1-2c$ and $a=c+4d+1=5-7c$ So $a \equiv 5 \pmod {7}$ Now to prove that any such $a$ can be answer Take $r(x)=(1+2n)x-5n-2$ for some integer $n$ and $p(x)=(x^2-2x-1)( (1+2n)x-5n-2)-x+3$ Then $p(\sqrt{2}+1)=2-\sqrt{2}$ and $p(\sqrt{2}+2)=7n+5$