Solve it!

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Phlembac Adib Hasan
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Solve it!

Unread post by Phlembac Adib Hasan » Mon Jun 04, 2012 11:56 pm

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

Unread post by Phlembac Adib Hasan » Mon Jun 04, 2012 11:58 pm

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

Unread post by Fahim Shahriar » Fri Jun 08, 2012 11:19 am

x=0 and y=0
Name: Fahim Shahriar Shakkhor
Notre Dame College

sakibtanvir
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Re: Solve it!

Unread post by sakibtanvir » Fri Jun 08, 2012 5:03 pm

My approach:
$(y+1)(y-1)=2^{x}-2=2(2^{x-1}-1)$
And with the equation,the problem is killed by crossfire.... :lol:
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

mutasimmim
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Re: Solve it!

Unread post by mutasimmim » Sat Nov 17, 2012 7:36 pm

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).

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harrypham
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Re: Solve it!

Unread post by harrypham » Sat Aug 10, 2013 7:21 am

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 $x=0$.

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