I myself can get the job done

That was a really interesting question!

Solution:

Using division algorithm for polynomials we have:

$f(x)=K(x)Q(x)+R(x)$

Plugging $K(x)=x^2 -1$ we have ,

$f(x)=(x^2-1)Q(x) + R(x)$

Now if for some $a$ (which I don't care is an imaginary number or not) we had $a^2 -1=0$ then $f(a)=R(a)$

So simply plug $x^2=-1$ in $f(x)$ to get the desired result:

$f(x)=7(x^2)^{16} + 5(x^2)^{11} + 3(x^2)^6 +(x^2) = 7 - 5 +3 -1 =4$

$x^{ 2 }+1\quad divides\quad x^{ 10 }+1.\\ 7x^{ 32 }+5x^{ 22 }+3x^{ 12 }+x^{ 2 }=4x^{ 32 }+(x^{ 10 }+1)\times k(x).\\ So,\quad we\quad only\quad have\quad to\quad find\quad remainder\\ when\quad 4x^{ 32 }\quad is\quad divided\quad by\quad x^{ 2 }+1.\\ \\ This\quad is\quad a\quad fairly\quad easy\quad process.\quad Follow\quad the\quad pattern\\ and\quad you\quad get\quad upto\quad -4\quad in\quad the\quad quotient.\quad So,\quad you\quad shoud\\ get\quad -(-4)\quad after\quad subtraction\quad and\quad you\quad end\quad up\quad with\quad 4\\ as\quad the\quad remainder.$

x becomes (i)... now if the power of x is divisible by 4... its positive else its negative......... therefore it come 7-5+3-1