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

Question

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?

My task is to:

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!

It's not immediately clear how a proof by cases would help, besides, perhaps, considering $x,y$ each modulo some pithy number. I'm voting to close this as off-topic as it is not evident that you have tried the problem yourself. — Shaun, Feb 05 '19 at 15:46

amWhy · Accepted Answer · 2013-02-15T14:33:36.423

10

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

edited Feb 15 '13 at 14:33

answered Feb 14 '13 at 20:13

amWhy

209,954

That would be considered proof by cases wouldn't it? :S – choloboy Feb 14 '13 at 20:14
1

Yes, it would: consider all scenarios: prove in each case, no solution exists. – amWhy Feb 14 '13 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 '13 at 20:19
3

@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

score 1 · Answer 2 · answered Feb 14 '13 at 20:13

1

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

answered Feb 14 '13 at 20:13

Valtteri

3,154

What if $x < 0$? Also, 1 is not even. – Emily 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

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

2 Answers2