# Find all integers $(x, y)$ such that $1 + 2^x + 2^{2x + 1} = y^2$ [duplicate]

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.

• artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/… Commented Jan 25, 2017 at 18:54
• Maybe useful$$1 + 2^x + 2^{2x + 1} = y^2\\2.2^{2x}+2^x+1-y^2=0 \\like\\ax^2+bx+c=0\\2^x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4(2)(1-y^2)}}{4}\\ \Delta=8y^2-7=q^2$$ Commented Jan 25, 2017 at 19:19
• 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 Commented Jan 25, 2017 at 19:31
• 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 $2$. Commented Jan 25, 2017 at 19:31

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

$x<-1$ 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.

• can you please explain me the last part of your proof starting from: Clearly we cannot have $k=2^{x-2}$ or $k+1=2^{x-2}$... Thanks in advance. Commented Jun 21, 2019 at 18:22
• @Kamal : First case implies $8k+1=k+1$, the second $8k+9=k$, which clearly are both not possible. Both are conclusions from $[2^{x-2}](8\cdot [2^{x-2}]+1)=k(k+1)$. Commented Jun 21, 2019 at 19:34
• @Lutz, thanks for giving me time to reply. I have one more question if this does not bother you. why we split $(2^{x+1}+1)$ into odd factors such that $2^{x-2}r = s\pm 1$ with r must be $|r|<8$? I did not get how he came to the last expression. thanks again. Commented Jun 22, 2019 at 8:18
• @Kamal : Generally, you split both factors, $ab=(a_1a_2)(b_1b_2)=(a_1b_1)(a_2b_2)$. However, from the first factor you get only powers of $2$, and one of the resulting factors has to be odd, so any finer factorization has to come from the second factor. Now you want to identify $(k,k+1)$ either with $(ab_1,b_2)$ or with $(b_2,ab_1)$, which requires $|b_2-ab_1|=1$. Commented Jun 22, 2019 at 8:37
• thanks a lot @LutzL for your beautiful explanation. Commented Jun 22, 2019 at 9:49