Basic Problems About Congruence 1

${ 6 }^{ x }+{ 8 }^{ x }\quad =\quad (7+1)^{ x }+(7-1)^{ x }\\ \therefore \quad f(77)=(7+1)^{ 77 }+(7-1)^{ 77 }\\ this\quad on\quad expansion\quad leaves\quad a\quad (multiple\quad of\quad 49)+(77*7)+(77*7)\\ this\quad can\quad be\quad written\quad as\quad (multiple\quad of\quad 49)+(49*22)\\ thus\quad f(77)\quad is\quad divisivle\quad by\quad 49\\ hence\quad remainder\quad is\quad 0.$

Firstly note that $6^x + 8^x = 2^x (3^x + 4^x)$ . With this, we need to evaluate the remainders of $2^{77}, 3^{77}, 4^{77}$ when divided by $49$ .

We have $2^{10} \equiv -5 \mod 49$ $2^{70} \equiv 30 \mod 49$ $\color{#D61F06}{2^{77} \equiv 18} \mod 49$ $2^{154} = \color{#3D99F6}{4^{77} \equiv 30} \mod 49$ $3^{10} \equiv 4 \mod 49$ $3^{70} \equiv 18 \mod 49$ $\color{#20A900}{3^{77} \equiv 19 \mod 49}$

Thus $6^{77}+ 8^{77} \equiv \color{#D61F06}{18} (\color{#20A900}{30} + \color{#3D99F6}{19}) \mod 49 \Leftrightarrow \boxed{6^{77}+ 8^{77} \equiv 0} \mod 49$ .

Any number of the form a^n + b^n/c, where a,b,c are in A.P, then they are divisible by a+b or c.