Solving A Homemade Diophantine Equation

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#DiophantineEquations #NumberTheoryProblems

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[stevenlitvintchouk]

I started off the same way as the video, by figuring that k = 0 to give us an odd integer. So 2^m - 2^n = 96. But 96 = 3 * 2^5. Divide both sides by 2^5. 2^(m-5) - 2^(n-5) = 3. And here we go again: 3 is an odd integer, but the only power of two that is odd is 2^0 = 1. So n-5 = 0, so n = 5. Therefore 2^(m-5) = 4, so m = 7. So 2^m - 2^n -2^k = 2^7 - 2^5 - 2^0 = 95.

[highlyeducatedtrucker]

Without even watching the video or reading any of the other comments or attempting to solve the question myself, I can state with a fair degree of confidence that k=0.

[Alex-kphd]

I solved the problem using binary




if m>7, then 1000...00 (m zeros) - would have more than 2 ones (its easy to prove by induction), so the number cannot be represented as a sum of two powers of 2, neither can m be less then 6, therefore m=7, the first one is in the 0th position, so n=0, and the second one is in the 5th position, so k=5.

[akatasis]

Solving 2^m - 2^n = 96 we have that m > n. Then m ≥ 7 (for non-negative answer) and m ≤ 7 (because the difference 2^m - 2^n must be less than 128). So m = 7 and n = 5.

[PunmasterSTP]

Diophantine? More like "I'm havin a fun time"...watching your videos. Thanks for making and sharing them!

[WahranRai]

Plagiat and copy paste forbidden !
Here is how to demonstrate that k= 0 → assume m > n > k (expression symmetrical in n and k) →
Let m = k+x and n= k+y → Substitute in the given expression and factorise → 2^k(2^x -2^y -1) = 95
The 2 factors must be odd → 2^k is odd only when 2^k = 1→ k=0 →2^x -2^y -1 = 95 →2^x = 96 +2^y = 2^6 + 2^5 + 2^y →
x > 6 → Try with x= 7 →2^7 = 128 = 96 + 32 = 96 + 2^5 → solution is : m = 7, n= 5 and k= 0

[Mutual_Information]

Diophantine equations are deceptive simple.. but can be so damn hard.

[rickostidich]

I usually think in hexadecimal (which is a very practical replacement for binary), and I write the quantity 96 as 60h (where "h" is a suffix and not a variable name). In hex, all the powers of 2 are numbers where a single digit is from {1, 2, 4, 8} and all the other digits are zeroes. Examples: 400h, 0.08h. Obviously, 60h=80h-20h, and 5Fh=80h-20h-1.

[gdtargetvn]

+ We can easily see that m, n and k must be non-negative. Otherwise, LHS is not integer, contrary to 95 being an integer.
+ Since 95 is odd, only one in these number: 2^m, 2^n or 2^k has to be odd.
+ Moreover, 2^m > 2^n + 2^k (since 95 is positive), so m > n, k ≥ 0 and either 2^n or 2^k has to be odd.
+ WLOG, supposing 2^k is odd => 2^k = 1 => k = 0
=> 2^m - 2^n = 96
=> 2^n.(2^(m-n)-1) = 96 (since m > n)
+ Also notice that 2^(m-n)-1 is always odd for all m>n, and 96 = 32.3. Hence:
2^n = 32 and 2^(m-n)-1=3
=> n = 5 and m = 7

In conclusion, the solutions are (7; 5; 0) and (7; 0; 5).

[raunakkumar]

m>n>k
hence 2^k (2^m-k - 2^n-k -1)= 95
since 2 doesn't divide 95, there are no even factors of 2, hence 2 shouldn't divide the LHS either, hence k=0
2^m-2^n = 96
2^n(2^m-n -1) = 96 = 2^5*3
n= 5,
comparing
2^m-5= 4, m = 7
Hence (m, n, k) = (7, 5, 0)

[ruchikarfacts]

2^m - 2^n - 2^k = 95
Where : m greater than n & k
And
Either n greater than k or k greater than n
Let's come to the given equation
2^k{2^(m-k) - 2^(n-k) - 1} = 95
Compare both side
(1) 2^k = 1
k = 0
(2) 2^m - 2^n - 1 = 95
2^m - 2^n = 96
2^n{2^(m-n) - 1} = 2^5(3)
Compare both side
n = 5
&
2^(m-n) = 4
m-n = 2
m = 7
{m, n, k} = {7, 5, 0}
Now we can also right the given equation
2^m - 2^k - 2^n =95
In the same way we can solve
n = 0
k = 5
m = 7
{m, n, k} = {7, 0, 5}
m, n & k values are
{7, 5, 0} & {7, 0, 5}

[owlsmath]

It looked scary at first but as soon as you see k = 0 you're all set and it becomes easy.

[klausg]

Concerning your question if there are negative solutions: since m >0, 2^n +2^k = (2^(n-k) +1)• 2^k is an integer. Suppose that n>k: then n-k>0, so 2^k is an integer, ie k >0. And so is n >0. So no negative solutions if n>k. If n=k then 2•2^k is an integer so k=-1 is possible. And then n=-1; so 2^m=96 which is impossible. BUT REMARK: if we had the following problem:
2^m -2^n -2^k =127 we would have the solutions (8, 7, 0) (8, 0, 7) and (7, -1, -1).

[angelmendez-rivera]

The smallest power of 2 larger than 95 is 128 = 2^7. 128 – 95 = 33, and notice that 33 = 32 + 1, so 128 – 95 = 32 + 1, which is equivalent to 128 – 32 – 1 = 95, and notice that 128 = 2^7, 32 = 2^5, and 1 = 2^0. Therefore, m = 7, n = 5, k = 1. The fundamental theorem of arithmetic guarantees this solution must, if it exists, must be unique. The way that works is that if 2^m – 2^n – 2^k = 2^m' – 2^n' – 2^k', then you can factor both sides, and the fundamental theorem of arithmetic will force you to conclude that m = m', and n = n' or n = k'. Therefore, the unique solution to this equation is (k, m, n) = (0, 7, 5).

[okim]

式がわかれば小学生でも解ける問題かな？



多くのm, n, k で項が偶数になるところ、右辺が奇数なので普通に考えてm, n,kのうちどれかが0で奇数を作る必要がある。
全て奇数もありだが、全て奇数(m=0, n=0, k=0)だと左辺が -1 になってしまうので不適。
m=0の場合、他２項が負になり左辺の計が負になるので不適。
nとkは対称なのでk=0と置く
2^m-2^n-2^0=95
2^m-2^n-=96
2^m-2^n-=3*32
2^m-2^n-=(4-1)*32
2^m-2^n-=4*32-1*32

[tcoren]

In binary We want to add two powers of 10 (2) to it such that it will result in a power of 10. The answer is so 2^0, 2^5 to get 2^7

[williamhogrider]

It's fun to see it done algebraically 🍺🍺🍻.

[DilipKumar-nskl]

It is a oneline soln.
2^m=2^n+2^p+95
m=7, n=0, p=5
m=7, n=5, p=0
Two solns.

[davidseed]

at 5:32. 2^p-1 must be odd.
so the factor pair must be 32x3

[TedHopp]

@5:35 you don't have to start with all that testing whether the second factor is one less than a power of 2. All you have to do is notice that 2^t-1 is odd, so the second factor must be odd. The only one on your list is 3. Done!