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

In summary: So 2^n remind 1 if n is 3+4k. So we have to find n if it is 3+4k and also 3+4k if n is 3+4k.In summary, the conversation discusses the equation (22016 + 5)m + 22015 = 2n + 1 and attempts to find all possible pairs of positive integers (m,n) that satisfy it. The solution is found to be only one pair, m = 1
  • #1
giokrutoi
128
1

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 ?
 
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  • #2
giokrutoi said:
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.
 
  • #3
haruspex said:
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
 
  • #4
giokrutoi said:
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?
 
  • #5
(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

 
  • #6
giokrutoi said:
24 is not 2^n
Right, but how do you know that a2+ 10a + 24 + a/2 is not of the form 2^n?
 
  • #7
a(a + 10 + 1/2 ) + 24
24 has only 2^3 times 3
so 22106 can't be equalized by 2^3
 
  • #8
giokrutoi said:
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 5m-1 cannot have a factor that is some large power of 2.
 
  • #9
haruspex said:
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 5m-1 cannot have a factor that is some large power of 2.
can it be done by induction?
 
  • #10
giokrutoi said:
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)?
 
  • #11
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
 
  • #12
and also if i pretend graphic of those functions they can't be equal
 
  • #13
giokrutoi said:
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?
 
  • #14
haruspex said:
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.
 
  • #15
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 ??
 
  • #16
Speaking about ends

22016 ends with 6
22015 ends with 8
Thus
(22016 + 5)m + 22015 ends with 9
n = 4k + 3
 
  • #17
giokrutoi said:
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.
 
  • #18
haruspex said:
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
 
  • #19
giokrutoi said:
i really don't understand what can be done that way
(22016+ 5)m=22016m(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?
 
  • #20
haruspex said:
binomial theorem
whats that?
 
  • #22
Here's an illustration of the method I'm suggesting, but applied to a trivial problem: 2m+1=2n.
The answer is obvious, and you could prove it using divisibility arguments, but here's how it would go with magnitude arguments...
2m<2n
m<n
m+1 <= n
2m+1<=2n=2m+1
2m<=1
m<=0
See if you can do something similar with the problem in this thread.
 
  • #23
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.
 
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  • #24
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
 
  • #25
giokrutoi said:
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.
 
Last edited:
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  • #26
haruspex said:
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
 
  • #27
giokrutoi said:
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.
 

1. What is the purpose of finding all pairs of positive integers for the equation (22016 + 5)m + 22015 = 2n + 1?

The purpose of finding all pairs of positive integers for this equation is to solve for the values of m and n that satisfy the equation. This can help us better understand the properties and patterns of positive integers and their relationship to each other.

2. How can this equation be solved to find all possible pairs of positive integers?

This equation can be solved by using algebraic methods such as substitution, elimination, or graphing. We can also use number theory concepts such as prime factorization and divisibility rules to narrow down the possible solutions.

3. Are there any restrictions on the values of m and n in this equation?

Yes, both m and n must be positive integers. This means they must be whole numbers greater than zero. Additionally, n must be an odd number since the right side of the equation is always odd (2n + 1).

4. Can there be more than one pair of positive integers that satisfy this equation?

Yes, there can be multiple pairs of positive integers that satisfy this equation. In fact, there are infinitely many pairs of positive integers that satisfy this equation since m and n can take on any value as long as they are positive integers.

5. How can finding all pairs of positive integers for this equation be useful in real-world applications?

Finding all pairs of positive integers for this equation can have applications in fields such as cryptography and number theory. It can also help in understanding the properties of positive integers and their relationships to other numbers. Additionally, this equation may have connections to other mathematical concepts and problems, making it a valuable tool for further exploration.

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