Exponential Diophantine Equation

We can write the equation $6n+4={ 2 }^{ i }$ as $6n={ 2 }^{ i }-4\Longrightarrow 3n={ 2 }^{ i-1 }-2$ . We can see that as ${ 2 }^{ i-1 }-2$ is even, then $3n$ is even; so let say $n=2{ n }_{ 1 }\Longrightarrow 6{ n }_{ 1 }+2={ 2 }^{ i-1 }$ which is $3{ n }_{ 1 }+1={ 2 }^{ i-2 }$ .

Now, the last equation says ${ 2 }^{ i-2 }\equiv 1(mod\quad 3)$ , this can be written as $(-1)(-1)...(-1)\equiv 1(mod\quad 3)$ with $-1$ written $i-2$ times (because $2\equiv -1(mod\quad 3)$ ). So, we can see that $i-2$ must be even.

With, this we suppose the first solutions are $i-2=2,4,6,8,10,$ but the solutions for non-negative $n$ are $i-2=4,6,8,10,12\Longrightarrow i=2,4,6,8,10$

So, when we replacethe values of $i$ in the first equation; we know that the first 6 $n$ solutions are:

$n=0,2,10,42,170,682$ . Their sum is $\boxed { 906 }$ .

Rewrite the equation as $3n = 2^{i-1}-2$ .

We see that $2^{i-1} \equiv 2 \pmod{3}$ .

From this, multiplying both sides by two gives $2^i \equiv 4 \equiv 1 \pmod{3}$ . By Fermat's little theorem, $i = 2k$ .

Hence, the sum of values are $\displaystyle \sum_{k=1}^6 \frac{2^{2k}-4}{6} = \boxed{906}$ .

6n+4= 2 ^ i

put

n=0 we get 4=2^i => i =2

n=2 6(2)+4= 2^i => 16= 2^i = > i=4

n=10 6(10)+4= 2^i =>64=2^i => i =6
by putting values of n we have arithmetic seq. of range 2, 4 , 6 with common difference d=2 so next values of n can be fouund by puuting values of i.

if i=8 then 6n+4= 256 => n= 42
similarly i= 10 6n+4= 1024 => n=170

i=12 6n+4 = 4096 => n= 682
now 0+2+10+42+170+682= 906