# Showing there are no integer solution to equation $\;2^x = 4y+3$

I am stuck on this problem and I'm not sure how to approach it. Can anyone help me out with figuring how to approach the proof?

Prove that it is impossible to find integers $\,x,\, y\,$ such that $\;2^x = 4y + 3$.

I assumed a proof by cases would be the way to go?

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Proof-By-Cases - Sketch:

We consider $x \in \mathbb{Z}$. For all $x \in \mathbb{Z}$:

1. $x > 0$
2. $x = 0,\;$ or
3. $x < 0$

$(1)$ 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: $$2^x = 4y + 3$$

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That would be considered proof by cases wouldn't it? :S –  choloboy Feb 14 '13 at 20:14
Yes, it would: consider all scenarios: prove in each case, no solution exists. –  amWhy Feb 14 '13 at 20:17
@theolc Not really. You can write the proof as: $4y+3$ is odd thus $2^x$ is odd integer, thus $x=0$ contradiction.... –  N. S. Feb 14 '13 at 20:19
@N.S.: The proof as sketched by amWhy is by cases; what you mean is that it can be reorganized to have a different structure. –  Brian M. Scott Feb 14 '13 at 22:40
The Op edited the question strangely. +1 –  B. S. Feb 15 '13 at 8:24
$2^x$ is even and $4 y+3$ is odd...
What if $x < 0$? Also, 1 is not even. –  Arkamis Feb 14 '13 at 20:25
@Arkamis If $x<0$, $2^x$ is not an integer and if $x=0$, it is one and the other side cannot be one... –  Valtteri Feb 14 '13 at 20:29