Find all nonnegative integer $x,y$ such that $y^2+1=2^x$
(It's a mathlinks problem!I solved yesterday.)
Solve it!
 Phlembac Adib Hasan
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Solve it!
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 Phlembac Adib Hasan
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Re: Solve it!
My favorite killing trick finishes this just in one line:
http://www.matholympiad.org.bd/forum/vi ... =14&t=1911
http://www.matholympiad.org.bd/forum/vi ... =14&t=1911
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 Fahim Shahriar
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Re: Solve it!
My approach:
$(y+1)(y1)=2^{x}2=2(2^{x1}1)$
And with the equation,the problem is killed by crossfire....
$(y+1)(y1)=2^{x}2=2(2^{x1}1)$
And with the equation,the problem is killed by crossfire....
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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Re: Solve it!
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).
Re: Solve it!
If $x \ge 2$ then $4y^2+1$. Therefore $y^2 \equiv 3 \pmod{4}$, a contradiction.Phlembac Adib Hasan wrote:Find all nonnegative integer $x,y$ such that $y^2+1=2^x$
(It's a mathlinks problem!I solved yesterday.)
Thus, $x=1$ or $x=0$.