Diophantine Equations Of The Power Dynasty

Given that $1 + 4^x + 4^y = z^2 \implies 4^x + 4^y = z^2 - 1 = (z-1)(z+1)$ . Since the LHS is divisible by $4$ , the RHS must also be divisible by $4$ . Either $z-1$ or $z+1$ is divisible by $4$ . Assuming $z-1$ is divisible by $4$ (it will be the same result if we assume $4 \mid z+1$ ), and let $z-1 = 4m$ , then:

$\begin{aligned} 4^x + 4^y & = (z-1)(z+1) \\ & = 4m(4m+2) \\ & = 4^2m^2 + 8m & \small \blue{\text{Since }x, y > 1 \text{, we can divide both sides by 4.}} \\ 4^{x-1} + 4^{y-1} & = 4m^2 + 2m & \small \blue{\text{For }4 \mid \text{RHS }\implies m = 2(4^n) \text{, where}} \\ & = 4^{2n+2} + 4^{n+1} & \small \blue{n \text{ is a non-negative integer.}} \end{aligned}$

Assumming $x < y$ , equating the exponents, we have $x = n + 2$ and $y=2n+3$ . And the solutions for $1 < x, y, z < 600$ are:

$\begin{array} {llll} n = 0 & \implies x = 2 & y = 3 & z = 9 \\ n = 1 & \implies x = 3 & y = 5 & z = 33 \\ n = 2 & \implies x = 4 & y = 7 & z = 129 \\ n = 3 & \implies x = 5 & y = 9 & z = 513 \end{array}$

We will get $4$ more solutions swapping $x$ with $y$ . Therefore the total number of solutions is $\boxed 8$ .

Let's consider the case $y\ge x$ and put $y=x+k$ . Note that $z$ must be odd; so put $z=2a+1$ . The equation is now

$1+4^x+4^{x+k}=1+4a+4a^2$

Tidying up,

$4^{x-1} \left( 1+4^k \right) = a(a+1)$

One of $a$ and $a+1$ is odd, the other even, and they are consecutive integers. So we must have $a=4^{x-1}$ and $a+1=1+4^k$ (assigning these the other way round doesn't work, as powers of $4$ are too far apart).

From this, then, we get $x-1=k$ , or $x=k+1$ , $y=2k+1$ . We're told $x>1$ ; so $k\ge1$ (and, importantly for counting solutions later, $x$ and $y$ are distinct).

It's easy to check back through and see $z=1+2^{2k+1}$ .

Since we've shown all solutions are of this form, we just need to check the largest possible $k$ such that $z<600$ ; this is $k=4$ , giving $z=513$ . So there are $4$ solutions; but recall that we assumed $y \ge x$ , so we need to double this count to get a grand total of $\boxed8$ solutions.

I used the simple code below to answer this question, the computation can be reduced if we observe that $4^{10} > 600^2$ and that the left side of the equation is always an odd integer. Moreover, if $(x,y,z)$ is a triple satisfying the equation and $x$ and $y$ are distinct then $(y,x,z)$ will also be a solution. The following is the output: $(1,1,3) , (2,3,9) , (3,2,9) , (3,5,33) , (4,7,129) , (5,3,33) , (5,9,513) , (7,4,129) , (9,5,513)$ . Since $x,y,z>1$ there are $8$ possible triples.

int main(){

    int x,y,z;

for( x=1 ; x<10 ; x++ )
for( y=1 ; y<10 ; y++ )
for( z=1 ; z<600 ; z+=2 )
if( (1+pow(4,x)+pow(4,y)-z*z)==0 )
printf("(%d,%d,%d) \n",x,y,z);

return 0;

}