BdMO Online Forum

Find all non-negative integer $x,y$ such that $y^2+1=2^x$
(It's a mathlinks problem!I solved yesterday.)

My favorite killing trick finishes this just in one line:
http://www.matholympiad.org.bd/forum/vi ... =14&t=1911

x=0 and y=0

My approach:
$(y+1)(y-1)=2^{x}-2=2(2^{x-1}-1)$
And with the equation,the problem is killed by crossfire....

Taking x=0 gives y=0, again x=1 gives y=1.Clearly x=2 have no integer solution.Taking x is equal or greater than 3 gives the right side is congruent to 0(mod8) but the left side is congruent to 1+1=2 (mod8)[y must be even].So the only solutions are (x,y)=(0,0),(1,1).

Phlembac Adib Hasan wrote:Find all non-negative integer $x,y$ such that $y^2+1=2^x$
(It's a mathlinks problem!I solved yesterday.)

If $x \ge 2$ then $4|y^2+1$. Therefore $y^2 \equiv 3 \pmod{4}$, a contradiction.
Thus, $x=1$ or $x=0$.