Finding All Pairs of Positive Integers for (22016+ 5)m + 22015 = 2n + 1

[giokrutoi]

Homework Statement

(2²⁰¹⁶+ 5)^m + 2²⁰¹⁵ = 2ⁿ + 1[/B]

find every n and m pairs
as they are positive integers

The Attempt at a Solution

(2²⁰¹⁶+ 5)⁰ + 2²⁰¹⁵ = 2²⁰¹⁵ + 1[/B]
so one pair is m= 0 , n = 2015
if m =1 the equation is meaningless
if m> 1 so there are really amount of powers that can't be handled by n as 1 in the end of the equation is needed to cancel out 5^m so 5^m -1 can't be equal to 2^n

so the equation has only one answer
am i right ?

 

[haruspex]

5^m -1 can't be equal to 2^n

False, e.g. m=1, n=2.

Also, m=0, n=2015 is not a solution since it says positive integers, so 0 is not allowed.

 

[giokrutoi]

False, e.g. m=1, n=2.

Also, m=0, n=2015 is not a solution since it says positive integers, so 0 is not allowed.

sorry non negative integers so 0 is allowed

and about m =1 n = 2 that is the only one that is not true for that equation and i considered m = 1 specifically so m>1 is only valid

 

[haruspex]

sorry non negative integers so 0 is allowed

and about m =1 n = 2 that is the only one that is not true for that equation and i considered m = 1 specifically so m>1 is only valid

Ok, but I don't see in your post a proof that there is no solution for m>1. Can you explain in more detail?

 

[giokrutoi]

(2²⁰¹⁶+ 5)^m + 22015 = 2ⁿ + 1
2²⁰¹⁶ i'll call this "a"
if m = 2 a² + 10a + 24 + a/2 = 2^n
24 is not 2^n n is natural number there is no n
m = 3 a³ + 124 + 15a²+ 75 a + a /2 = 2^n
124 is not 2^n n is natural number there is no n
and as the m is increasing and all the power is increasing n won't be found

 

[haruspex]

24 is not 2^n

Right, but how do you know that a²+ 10a + 24 + a/2 is not of the form 2^n?

 

[giokrutoi]

a(a + 10 + 1/2 ) + 24
24 has only 2^3 times 3
so 2²¹⁰⁶ can't be equalized by 2^3

 

[haruspex]

a(a + 10 + 1/2 ) + 24
24 has only 2^3 times 3
so 2²¹⁰⁶ can't be equalized by 2^3

Ok, that deals with m=2, but I still don't see that it works for m in general. You would need to show that 5^m-1 cannot have a factor that is some large power of 2.

 

[giokrutoi]

Ok, that deals with m=2, but I still don't see that it works for m in general. You would need to show that 5^m-1 cannot have a factor that is some large power of 2.

can it be done by induction?

 

[haruspex]

can it be done by induction?

Maybe, but two usual approaches to this sort of problem are (what I call) modulo arithmetic; and magnitude analysis.
E.g., for large m, what roughly must n equal? Must it be a bit more or a bit less than that? Since it must be an integer, what are the consequences of its being a whole 1 more (or 1 less)?

 

[giokrutoi]

as 5 goes to some power 2 mast go power with much more than that maybe for 5 to 20 2 to 50 there are no possible for then to equal

 

[giokrutoi]

and also if i pretend graphic of those functions they can't be equal

 

[haruspex]

as 5 goes to some power 2 mast go power with much more than that maybe for 5 to 20 2 to 50 there are no possible for then to equal

You will need a far more rigorous argument than that!
Try pulling the 2²⁰¹⁶ term outside the parentheses, leaving (1+5/2²⁰¹⁶) inside. What can you then say about the approximate relationship between n and m?

 

[giokrutoi]

You will need a far more rigorous argument than that!
Try pulling the 2²⁰¹⁶ term outside the parentheses, leaving (1+5/2²⁰¹⁶) inside. What can you then say about the approximate relationship between n and m?

Hey I was thinking a lot so I found this out look 2^2016 + 5 decided by five in remind is 1 so the last number is or 1 or 6. And remind stays the same now 2^2015-1 divide by 5 and remind is 2 so on the right side where we have these numbers. Divide whole equation by 5 gives a us remind 3 now on the left side where we have 2^n this number has period 4. 2^1 / 5 remind is 2 2^2/5 remind 4. 2^3. Remind 3. 2^4 /5. Remind is 1. So if we want reminds to be equal each other we have to choose only. 2^3+4k. Where k is a natural number. This is the first thing.

 

[giokrutoi]

The second thing is this look if we divide 2^2016+5 by 3 remind is 0 and 2^2015 - 1 by 3 remind is 1. So 2^n devided by 3 must remind 1 . Period of 2^n is 4 so if the power of 2 is even remind after division by 3 is 1 and if power of 2 is odd after division remind will be 2
So the equation to be equal power n must be even but 2^3+4k is odd so odd can't be equal even so there can't be found n and m .

Are my conclusion right ??

 

[Irene Kaminkowa]

Speaking about ends

2²⁰¹⁶ ends with 6
2²⁰¹⁵ ends with 8
Thus
(2²⁰¹⁶ + 5)^m + 2²⁰¹⁵ ends with 9
n = 4k + 3

 

[haruspex]

The second thing is this look if we divide 2^2016+5 by 3 remind is 0 and 2^2015 - 1 by 3 remind is 1. So 2^n devided by 3 must remind 1 . Period of 2^n is 4 so if the power of 2 is even remind after division by 3 is 1 and if power of 2 is odd after division remind will be 2
So the equation to be equal power n must be even but 2^3+4k is odd so odd can't be equal even so there can't be found n and m .

Are my conclusion right ??

You can certainly conclude many facts about what n must be modulo various bases, but I don't see any of these lines leading to m=0 being the only solution. (2^3+4k is even.)

The only way I can see to obtain the desired answer is the magnitudes method I mentioned in post #10 and laid out the first steps for in post #13.

 

[giokrutoi]

You can certainly conclude many facts about what n must be modulo various bases, but I don't see any of these lines leading to m=0 being the only solution. (2^3+4k is even.)

The only way I can see to obtain the desired answer is the magnitudes method I mentioned in post #10 and laid out the first steps for in post #13.

i really don't understand what can be done that way

 

[haruspex]

i really don't understand what can be done that way

(2²⁰¹⁶+ 5)^m=2^2016m(1+5.2^-2016)^m
Now apply the binomial theorem to expand the part in parentheses. Try keeping the first three terms, and we'll see how it goes. If you know the 'big O' notation you can use that for the third and subsequent terms.
Based on this, what can you say immediately about n as a function of m, asymptotically?

 

[giokrutoi]

binomial theorem

whats that?

 

[haruspex]

whats that?

It's pretty basic. Perhaps you know it by another name.
See https://en.m.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem and the first item under https://en.m.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Characterizations.

 

[haruspex]

Here's an illustration of the method I'm suggesting, but applied to a trivial problem: 2^m+1=2ⁿ.
The answer is obvious, and you could prove it using divisibility arguments, but here's how it would go with magnitude arguments...
2^m<2ⁿ
m<n
m+1 <= n
2^m+1<=2ⁿ=2^m+1
2^m<=1
m<=0
See if you can do something similar with the problem in this thread.

 

[haruspex]

A couple of things I should make clear:
- I do not know whether there are any other solutions (do you know whether there are supposed to be?)
- Even if there are not, I do not know whether my suggested magnitudes argument works here. I just know that it can be useful.
That said, unless this is at a fairly advanced level, I suspect that there are no other solutions.

Edit: I think I can show that if m>0 then m>2²⁰⁰⁰, or something like that.

 

[giokrutoi]

i tried your way but i couldn't do that kind of things
another way of solution are reminds that i proposed i don't have another

 

[haruspex]

i tried your way but i couldn't do that kind of things
another way of solution are reminds that i proposed i don't have another

Here's where I got to:
$2^{2016m}(1+5.2^{-2016})^m+2^{2015}=2^n+1$
Clearly n > 2016m, so n>= 2016m+1
$2^{2016m}(1+5.2^{-2016})^m+2^{2015}>=2^{2016m+1}$
Now, for x, r > 0, $(1+x)^r<e^{rx}$
$2^{2016m}e^{5m2^{-2016}}+2^{2015}>2^{2016m+1}$
$e^{5m2^{-2016}}+2^{2015-2016m}>2$
If m>0 then the second term less than 1/2:
$e^{5m2^{-2016}}>3/2$
$m>\frac 15 2^{2016}\ln(\frac 32)$
It looks like it should possible to cycle that back into the first equation above to show that n>=2016m+2. Might be a proof of no solutions along those lines.

 

[giokrutoi]

Here's where I got to:
$2^{2016m}(1+5.2^{-2016})^m+2^{2015}=2^n+1$
Clearly n > 2016m, so n>= 2016m+1
$2^{2016m}(1+5.2^{-2016})^m+2^{2015}>=2^{2016m+1}$
Now, for x, r > 0, $(1+x)^r<e^{rx}$
$2^{2016m}e^{5m2^{-2016}}+2^{2015}>2^{2016m+1}$
$e^{5m2^{-2016}}+2^{2015-2016m}>2$
If m>0 then the second term less than 1/2:
$e^{5m2^{-2016}}>3/2$
$m>\frac 15 2^{2016}\ln(\frac 32)$
It looks like it should possible to cycle that back into the first equation above to show that n>=2016m+2. Might be a proof of no solutions along those lines.

thank you very much i didn't knew the formula for (1+x)^r
that is really good proof

 

[haruspex]

thank you very much i didn't knew the formula for (1+x)^r
that is really good proof

Standard binomial expansion:
$(1+x)^r=1+rx+\frac 12r(r-1)x^2+\frac 16r(r-1)(r-2)x^3+... +{r\choose k}x^k+...$
And for exponentials:
$e^y=1+y+\frac 12y^2+\frac 16y^3+...+\frac 1{k!}y^k+...$
By inspection, putting y=rx, each term in the first series is less than or equal to the corresponding term in the second series.