If $$x,y$$ are positive integers such that $$2^x+49 = y^2$$,
find the value of $$x$$ and $$y$$.

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Uniteasy Admin · Accepted Answer

Since $$2^x$$ are multiples of $$2$$, $$2^x$$ is even. Addition of even and odd numbers results in an odd number. Therefore, $$y^2$$ is odd , so it is for $$y$$
$$2^x+49 = y^2$$n
$$2^x = y^2-49=(y-7)(y+7)$$n
So $$y-7$$ and $$y+7$$ are numbers of power of 2
Let 
$$y-7 = 2^k$$n
$$y+7 = 2^m$$n
Then
$$2y = 2^k+2^m =2^k(1+2^{m-k})$$
$$y = 2^{k-1}(1+2^{m-k})$$
To make $$y $$be odd, the only possible is to make
$$k-1 = 0$$, that is
$$k =1$$
Substitutng to (3) gives
$$y = 7+2 = 9 $$
Substitute y to (2) 
$$2^x = (9-7)(9+7) = 32$$
$$x = 5 $$

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