Showing there are no integer solution to equation $\;2^x = 4y+3$
Proof-By-Cases - Sketch:
We consider $x \in \mathbb{Z}$. For all $x \in \mathbb{Z}$:
- $x > 0$
- $x = 0,\;$ or
- $x < 0$
$(1)$ For non-negative integer $x (x >0)$: Show the left hand side will always be even, except when $x = 0$, and the right hand side will always be odd, regardless of the integer value of $y$. (I.e. all positive integral powers of $2$ are even, but $4y+3 = 2\cdot 2 y + 2 + 1 = 2(2y+1) + 1$ must be odd, regardless of the value of $y$.)
$(2)$ Then consider the case $x = 0$: $\;2^0 = 1 \neq 4y+3 = 2(2y+1) + 1$, whatever the integer value of $y$.
$(3)$ For negative integers $x (x < 0):$ the left-hand side will not be an integer $\left(\text{e.g.,}\;\; 2^{-2} = \dfrac 14\right),\;$ while the right hand side will always be an integer, regardless of the value of integer $y$. Hence the equation has not solution in integers in this case, either.
And hence we conclude there are no integer solutions for $x, y$ satisfying the equation: $$2^x = 4y + 3$$