What comes to your mind when you look at powers of 2?

Number Theory Level 3

$\Large{(x+y)(1+xy) = 2^z}$

How many non-ordered pairs of non-negative integers $x,y,z$ exists satisfying the above equation?

Data Insufficient. Can't be solved. 5

\large{\infty}

2 0 4 1 3

2 solutions

Abdeslem Smahi
Aug 9, 2015

If we consider $x=2^k-1$ and $y=1$ then $(x+y)(1+xy)=2^{2k} \implies z=2k$

which means some of the solutions are $(x,y,z)=(2^k,1,2k)$

so there is infinitely many solution

Look at this : $(x,y,z) = (2^k - 1, 2^k +1, 3k+1)$ ?

Satyajit Mohanty - 5 years, 10 months ago

better than mine , love it :)

Abdeslem Smahi - 5 years, 10 months ago

Nelson Mandela
Aug 9, 2015

If we consider y to be zero then,

(x)(1+0)=2^z.

Now for various integral values of x(positive powers of 2), we get infinite solutions.

(2,0,1);(4,0,2);(8,0,3)............................. infinite solutions.

Even (1,1,2) will satisfy the condition.