A Diophantine Equation with 2s | Number Theory

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[sanamite]

1) Odd, so n or k is 0, let's say k = 0 for the next step
2) 2^m - 2^n = 96 = 2⁵•3 = 2⁵(2²-2⁰) = 2⁷-2⁵
3) you get k or n is 0 and the other is 5 :
[7;5;0] or [7;0;5]

[mcwulf]

Obvs m > n, k
Assume n >= k
Factor out 2^k:
2^k (2^m-k - 2^m-n) = 95
Obvs k=0 for RHS to be odd.

So 2^m - 2^n = 96
2^n (2^m-n - 1) = 96 = 32 * 3
2^n = 32
n = 5

2^m-n - 1 = 3
2^m - n = 4
m - n = 2
m = 2 + 5 = 7

[allanmarder]

Clearly m> n and m>k or else the equation would result in a negative answer. Assume k is the smallest of the integers.
Then (2^k)*[2^(m-k) + 2^(n-k) -1] =95 Now 95 can be factored as 5*19 or 1*95 and one of the factors has to be a power of 2.
Thus the only possibility is for k=0 and 2^k=1. Then the second factor is (2^m -2^n -1)=95 or (2^n)*[2^(m-n)-1]=96. We can factor
96 as 2x48 4x24 8x12 16x6 32x3 96x1, The first factor must be a power of 2 and the second factor must be 1 less than a power of 2.
The only possibility is for the second factor to be 3 (2^2-1) and the factorization of 96 must be 32x3 so n=5 and m-n=2 or m=7.
Thus k=0 n=5 m=7. If we had assumed n to be the smallest of the integers we would get k=5 n=0 m=7.

[runenorderhaug]

Desmos 3d depection of it is interesting

[nasrullahhusnan]

Let p in an integer. 2^p =1 if p=0 and is even for p>0.
As RHS is odd, then either n or k, but not both, is 0. Without lost of generality, let n>k --> k=0
Thus (2^m)-(2^n)-1=95
(2^m)-(2^n)=96
=128-32
=2⁷-2⁵
Therefore (m, n, k)={(7, 5, 0), (7, 0, 5)}

[rdspam]

N or k must be 0. 2^0 is the only way to get an odd number. Then powers of 2 that are 96 apart must be 128 and 32. So 7, 5, and 0.

[neilmccoy]

a) k=0. b) lhs is multiple of 2^5 but not 2^6 ; so n=5 c) 2^m = 95+1+32=128=2^7. So, m=7

[phill]

Did it in binary - 96 -> Rounded up to to get n=7, subtracted to get to get m=5

[FisicTrapella]

There is another way to get an odd number, but it doesn't work. That is n = k = -1. (=> 2^m = 96; with no solution if m has to be an integer).