Let a and b be integer with
$$ a\ge 0\
b \ge 0\
2^n a + b \text { is a perfect square for any natural number } \, n \,

$$
Show that a=0.

Question

Let a and b be integer with 
$$ a\ge 0\
b \ge 0\
2^n a + b \text { is a perfect square for any natural number } \, n \, 

$$
Show that a=0.

anonymous · Answer

hello doc... :)

anonymous · Answer

Hi @dpalnc

anonymous · Answer

hello @eliassaab

shubhamsrg · Answer

is it correct? because for ,,let n=5 and a=1,,b=32 gives 64,,which is a perfect square..

anonymous · Answer

It is a perfect square for every n and not only for a particular n.

shubhamsrg · Answer

what i wanted to say is that when a not equals 0,,then also its a perfect square..

shubhamsrg · Answer

and the ques asks to prove a=0

myininaya · Answer

So it is $$2^na+b ?$$
or
$$2^{na+b}$$?

I just wanted to be sure you just meant to raise just the n

anonymous · Answer

$$
2^n \, a\, + b
$$

shubhamsrg · Answer

but as i said above,,when a not equals 0,,then also the eqn can be a perfect sq..

anonymous · Answer

You have to show that if a is not zero and 
$$
2^n \, a + b 
$$
is a perfect square for every n, then you must conclude a contradiction.

( for every n) and not only for a particular n.

shubhamsrg · Answer

ohh..not completely my fault though,,in the ques,,you wrote show a=0!! ;)

anonymous · Answer

But I said
"
is a perfect square for any natural number n
Show that a=0.

"

shubhamsrg · Answer

not getting it!! o.O
bah!1 let it be..seems like not my cup of tea..

anonymous · Answer

Clearly the method to use is proof by contradiction, find a contradiction for the assumption that $a>0$. From there, though, I am not yet sure where to go.

dumbcow · Answer

lets see if you look at cases where n is even then 2^n is itself a perfect square
when n is odd, then 2^n is not a perfect square but a,b must satisfy all n

n is even, let a=1, then b must be 0
n is odd, and a=1, b=0 --> result is not a perfect square

n is even, let a = 0, then b must be some constant perfect square
n is odd , a=0, b = constant  --> result perfect square

experimentx · Answer

Let n=1,a=1, b=2
we have 2^1*1 + 2 = 4, perfect square but a=1 != 0

Not sure if i am getting the question right!!!

anonymous · Answer

@dumbcow is correct, but that is only a contradiction for a=1, not for a>0. We need a contradiction for all a>0.

dumbcow · Answer

yeah i need to generalize it but thats the right idea i believe

anonymous · Answer

Yes, definitely right place to start. I think we can do a case-by-case with n=odd or n=even and a>0 to find a contradiction in general.

anonymous · Answer

I dont understand the question :/ 

What is the question asking?

anonymous · Answer

$2^na + b$  is perfect square only when $a=0$?

dumbcow · Answer

correct for all n

anonymous · Answer

can we do it by induction? assuming the $2^na + b$ to be prefect for $a \neq 0$.

experimentx · Answer

seems like i misunderstood question!!

anonymous · Answer

perfect square* 

For $n=1$, $a+b$ is perfect square.
For $n=k$, Let $2^ka + b$ be perfect square. 

Hmm I dont think it's right :/

experimentx · Answer

b has to be a perfect square ..

anonymous · Answer

Let $2^ka+b=q^2$. 

$$2^{k+1}a + b = 2^ka + 2^ka + b = q^2 + 2^ka$$

anonymous · Answer

Hmm what's the condition for Pythagorean triplet? but before that, I will need to show if $2^ka$ is perfect square or not for $a\neq 0$.

dumbcow · Answer

are you trying to show a contradiction when a != 0 by using induction ?

anonymous · Answer

yeah, but i am probably not getting anywhere.

anonymous · Answer

If $k$ is even, a has to be perfect square and if $k$ is odd then a must be odd powers of 2 for $2^ka$ to be perfect square.

anonymous · Answer

hmm for $2^ka + b = q^2$ to be valid, $2^ka$ and $b$ have to be perfect square, other than being pythagorean triplets.

For $k+1$, $2^{k+1}a + b$ can't be perfect square if $a\neq 0$. 
Because: "If $k$ is even, $a$ has to be perfect square and if $k$ is odd then $a$ must be odd powers of 2 for $2^ka$ to be perfect square."

anonymous · Answer

Is this good enough?

anonymous · Answer

No. Here a hint:
First show that b cannot be zero.
Then show that the hypothesis and a not zero  implies the existence of two integers u and v such that
$$
u^2 < v^2 < (u+1)^2
$$
Which is impossible.

kinggeorge · Answer

Here's the way I'm thinking about this. Suppose $b=0$, then either $2a$ or $4a$ is not a perfect square (for $a\neq0$). So let $b\neq0$. Also, at some point, we know that $$2^na\leq b< 2^{n+1}a$$If $b=2^na$ then $2^na+b=2^{n+1}a$ and we're back to square one. If $b\neq 2^{n}a$, then we have that$$2\cdot2^{n}a<2^na+b<3\cdot2^na$$If we write $u=2^na$ we have the relation $$2 u>u+b <3u$$At some points, this must be false since $u$ varies. So that leaves us with two possibilities. 
1. $b=0$. If true, $a=0$ by the first part.
2. $b\neq0$. If this is true, we want $u=2^na$ to not vary. Hence, $a=0$. 

QED

anonymous · Answer

Should  2u>u+b<3u be
2u<u+b<3u

anonymous · Answer

Let
$$
u_n^2 = 2^{n+2} + b, \quad b \ne 0,\, a\ne0\

$$
Choose n so that
$$
u_n > \frac 3 2 b \

(u_n+1)^2 = u_n^2 + 2u_n +1 > u_n^2 +3b =\2^{n+2}a + 4 b = 4( 2^n a + b) = v^2 > 
2^{n+2}a +  b= u_n^2

 
$$
if we put 
$$

u = u_n\
\text { then }\ 
(u+1)^2 > v^2 > u^2

$$
Contradiction.

kinggeorge · Answer

Yes, $2u>u+b<3u$ should be $2u<u+b<3u$. Thanks for catching that.