# Showing there is 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 solve the solution.

The question is:

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?

Any input? Thanks in advance!

-

choloboy - please ask a different question in a new post :-)  FYI: one possible solution, in integers, to $4^x = 8y$ is $x = 2, y = 2$, and to show a solution is possible, you need only find one solution! – amWhy Feb 15 at 13:39

## 2 Answers

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: $$2^x = 4y + 3$$

-
That would be considered proof by cases wouldn't it? :S – choloboy Feb 14 at 20:14

1
Yes, it would: consider all scenarios: prove in each case, no solution exists. – amWhy Feb 14 at 20:17

2
@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 at 20:19

2
@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 at 22:40
The Op edited the question strangely. +1 – Babak S. Feb 15 at 8:24
show 1 more comment

$2^x$ is even and $4 y+3$ is odd...

-

What if $x < 0$? Also, 1 is not even. – Arkamis Feb 14 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 at 20:29
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8881505131721497, "perplexity_flag": "head"}