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**Soroban** Hello, starvin-marvin!

We have: .$\displaystyle 4k + 1 \:=\:b^2$, for some integer $\displaystyle b$

. . Then: .$\displaystyle k \:=\:\frac{b^2-1}{4}\;\;{\color{blue}[1]}$

We see that: $\displaystyle b^2-1 \text{ is even }\quad\Rightarrow\quad b^2\text{ is odd }\quad\Rightarrow\quad b\text{ is odd}$

Let: $\displaystyle b \:=\:2c+1$

Then [1] becomes: .$\displaystyle k \:=\:\frac{(2c+1)^2-1}{4} \:=\:c^2 + c \quad\Rightarrow\quad k \:=\:c(c+1)$

Hence, $\displaystyle k$ must be *the product of two consecutive integers.*

Then: .$\displaystyle x \;=\;\frac{1 + \sqrt{1 + 4k}}{2} \;=\;\frac{1 + \sqrt{1+4c(c+1)}}{2} \;=\;\frac{1 + (2c+1)}{2} \;=\;c+1$, an integer.