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**Soroban** Hello, cassiopeia1289!

Assume there exist integers $\displaystyle x$ and $\displaystyle y$ that satisfy the equation.

We have: .$\displaystyle x^2 \:=\:4y+3 \quad\Rightarrow\quad y \:=\:\frac{x^2-3}{4}$

$\displaystyle x$ must either even or odd

If $\displaystyle x$ is even, then $\displaystyle x \:= \:2k$ for some integer $\displaystyle k.$

. . Then: .$\displaystyle y \;=\;\frac{(2k)^2-3}{4} \;=\;\frac{4k^2-3}{4}\;=\;k^2 -\frac{3}{4}$

Hence, $\displaystyle y$ cannot be an integer.

If $\displaystyle x$ is odd, then $\displaystyle x \:= \:2k+1$, for some integer $\displaystyle k.$

. . Then: .$\displaystyle y \;=\;\frac{(2k+1)^2 - 3}{4} \;=\;\frac{4k^2 + 4k - 2}{4} \;=\;k^2+k - \frac{1}{2} $

Hence, *y* cannot be an integer.

Therefore, no integer solution exists.