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Find all integers $$(x, y)$$ such that $$1 + 2^x + 2^{2x + 1} = y^2$$ So I basically used $$ f(x) = 1 + 2^x + 2^{2x + 1} = y^2$$ and created a table from 0 to 20. I got two pairs of integers: $$(0, \pm2)$$ and $$(4, \pm23)$$ I want to know if there's another method or if there are any other pairs of integers, or a generalisation.

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    https://www.artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_4 – 2017-01-25
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    Maybe useful$$1 + 2^x + 2^{2x + 1} = y^2\\2.2^{2x}+2^x+1-y^2=0 \\like\\ax^2+bx+c=0\\2^x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4(2)(1-y^2)}}{4}\\ \Delta=8y^2-7=q^2$$ – 2017-01-25
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    before anyone closes this without looking at the link above, while this problem does seem to come from the IMO in 2006 the linked page doesn't have a correct answer (only answer fragments that the page admits are faulty). As such, the OP seems justified in searching for a correct answer... Though it would be much better if the OP explained this himself/herself – 2017-01-25
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    This can be rewritten as $(2^{x+2}+1)^2-8y^2=-7$. So you want a solution to $u^2-8y^2=-7$ for $u$ one more than a power of $2$. – 2017-01-25
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    Possible duplicate of [Solve the following equation in positive integers $x$ and $y$](https://math.stackexchange.com/questions/808509/solve-the-following-equation-in-positive-integers-x-and-y) – 2018-04-07

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$1 + 2^x + 2^{2x + 1} = y^2 \\\implies 2^x(2^{x+1}+1) = (y+1)(y-1)$

$x<-1$ gives a non-integer LHS (no solutions)

$x=-1$ gives LHS $= 1$ with no solutions for $y$

$x=0$ gives LHS $= 3$ and $y=\pm 2$

For $x>0$, $y$ is odd so put $y=2k+1$ and $2^x(2^{x+1}+1) = (2k+2)(2k) = 4k(k+1)$ which is divisible by $8$ so $x\ge 3$ and $2^{x-2}(2^{x+1}+1) = k(k+1)$.

Clearly we cannot have $k=2^{x-2}$ or $k+1=2^{x-2}$ so we need $(2^{x+1}+1)$ to split into (odd) factors $r,s$ such that $2^{x-2}r = s\pm 1$.

Then $|r|<8$ otherwise $2^{x-2}|r| > (2^{x+1}+1)$. Also $2^{x-2}r^2 = sr\pm r$ so $|r|=3$ is the only viable choice and $2^{x-2}9 = (2^{x+1}+1) \pm 3$ gives $2^{x-2} = 4$ i.e. $x=4$ (and $y=\pm 23$) as the only other solution.

In summary: the only solutions $(x,y)$ are $(0,\pm 2)$ and $(4,\pm 23)$, those you found.