If $c$ is odd=$2k+1$(say),
(i)$c^2=(2k+1)^2=8\frac{k(k+1)}{2}+1=8y+1$ for some integer $y$.
$c^4=(c^2)^2=(8y+1)^2=64y^2+16y+1=1+16(y+4y^2)=1+16z$ for some integer $z$.
(ii) $c^4=(2k+1)^4=(2k)^4+ ^4C_1(2k)^3+ ^4C_2(2k)^2+ ^4C_3(2k)+ 1$
$=16k^4+32k^3+24k^2+8k+1≡8k^2+8k+1\pmod {16}=16\frac{k(k+1)}{2}+1≡1\pmod {16}$
(iii)we have already found $8\mid(c^2-1)$
Now $2\mid(c^2+1)$ as $c$ is odd, so, $8\cdot 2\mid(c^2-1)(c^2+1)=>16\mid(c^4-1)$
(iv) Using this, $\lambda(16)=\frac{\phi(16)}{2}$ as $16$ is a power$(≥3)$ of $2$, so $\lambda(16)=4=>c^4≡ 1\pmod {16}$ if $(c,16)=1$ i.e., if $c$ is odd.
So, in all the 4 ways we have proved, $c^4$ leaves remainder $1$ when divided by $16$ if $c$ is odd.
So, $a^4$, $b^4$ each will leave $1$ as remainder when divided by $16$