Exponential divisibility?

$7^a-3^b$ is even, so $a^4+b^2$ has to be even as well. Hence $a$ and $b$ have the same parity.

Consider the case when both $a$ and $b$ are odd.

$a^4+b^2\equiv 1+1=2\mod 4$

$7^a-3^b\equiv 7-3\equiv 0\mod 4$

So $a^4+b^2$ cannot be divisible by $7^a-3^b$

Hence, both $a$ and $b$ are even.

Let $a=2m$ and $b=2n$ . Then $7^a-3^b=7^{2m}-3^{2n}=\dfrac{7^m-3^n}{2}×2(7^m+3^n)$ and both factors are integers. So $2(7^m+3^n)|(7^a-3^b)$ and $(7^a-3^b)|(a^4+b^2)$ =2(8m^4+2n^2))

Therefore, $7^m+3^n\leq 8m^4+2n^2$ $\color{#D61F06}{\text{.....(1)}}$

It can be proved by induction that $8m^4<7^m$ for $m\geq 4$ , $2n^2<3^n$ for $n\geq 1$ and $2n^2+9\leq 3^n$ for $n\geq 3$ .

For $m\geq 4$ , we get $7^m+3^n>8m^4+2n^2$ which contradicts obtained inequation $\color{#D61F06}{\text{(1)}}$ .

So we have $3$ possible cases,

• m=1. Then $a=2$ and $8+2n^2\geq 7+3^n$ thus, $2n^2+1\geq 3^n$ . This is only possible when $n\leq 2$ . Checking manually, we get $n=2$ satisfies, hence $(2,4)$ is a solution.

• m=2. Then $a=4$ and a^4+b^2=256+4n^2\geq |7^4-3^b|=|49-3^n|(49+3^n). The smallest value of first factor is \(22 attained at $n=3$ , so $128+2n^2\geq 11(49+3^n)$ , which is impossible since $3^n>2n^2$ .

• m=3. Then $a=6$ and a^4+b^2=1296+4n^2\geq |7^6-3^b|=|343-3^n|(343+3^n). The smallest value of first factor is \(100 attained at $n=5$ , so $324+n^2\geq 25(343+3^n)$ , which is impossible since $3^n>2n^2$ .

Hence, the only possible ordered pair $(a,b)=(2,4)$ . So the answer is $2+4=\boxed{6}$

There exists one ordered pair that satisfies the condition: (2, 4). This produces a value of 32 for $a^{4}$ + $b^{2}$ and -32 for $7^{a}$ - $3^{b}$ .

-32 evenly divides 32, so (2, 4) is a valid pair. Therefore the summation of all elements of the ordered pairs is 6.

$Algorithem~~for ~~solving~~through~~a~~TI-83~Plus~~Calculator$

$\color{#D61F06}{1~~STO\rightarrow~a~~\downarrow .}...........\downarrow ~means ~ENTER.\\ 0~~STO\rightarrow~b~~\downarrow \\ b+1~~STO\rightarrow~b~~\downarrow ~~~{\Huge : }~~(a^4+b^2)\div (7^a~-~3*b)~~\downarrow \\ \text{Go-on pressing ~ENTRY....(2nd~and ~ENTER)~}\\ \text{till~you~get~an~integer~as~an~answer.} \\ or~else~stop~at~say~b=10\\ ~~~\\ \color{#D61F06}{Next~ for~ a=2:-}\\ 2~~STO\rightarrow~a~~\downarrow \\ 0~~STO\rightarrow~b~~\downarrow \\ ENTRY...as ~above~till~you~get~an~integer~as~an~answer.~~or ~b=10\\ You~will~get~-1~~at~~b=4~~and ~~a=2.\\ \text{Go\_on increasing a by one starting from b stored as 0. }$