2021/016) Find prime numbers x,y z such that $2^{x+1} + y^2 = z^2$

We are given

$2^{x+1} + y^2 = z^2$

Hence $2^{x+1}= z^2 - y^2 = (z+y)(z-y)$

Because product of (z-y) and (z+y) is power of 2 hence each of them is a power of 2. 

Let $z+y=2^a\cdots(1)$

$z-y= 2^b\cdots(2)$

Then $x+1 = a +b\cdots$

Adding (1) and (2) we get $2z=2^a+2^b=> z = 2^{b-1}(2^{a-b} + 1)$

And subtracting we get  $2y=2^a+2^b => y =  2^{b-1}(2^{a-b} -1)$

y and z cannot be prime unless b =1 so we have b = 1

From above 2 we get $y = 2^{a-1} -1$ and $z=2^{a-1} + 1$

Now taking $y, 2^{a-1}, z$ we have $2^{a-1}-1,2^{a-1},2^{a-1} + 1$ 

3 consecutive integers and so one of them is divisible b 3. As $2^{a-1}$ is not divisible by 3 so either 

$2^{a-1}-1$ or $2^{a-1}+1$ is. and it has to be 3 else it cannot be a prime. Now if $2^{a-1}+1$ is 3 then $2^{a-1} -1$ is 1 and not a prime. 

so $2^{a-1} -1 = y = 3$ giving a = 2, z = y + 2 = 5  and putting it in the given equation z = 3

so solution set $(x=3,y=3,z = 5)$