Solve the Mystery: Three Non-Negative Integers & Perfect Powers of 2

[arpitm08]

Question: There are three non-negative integers with the following property: If you multiply any two of the numbers and subtract the third number, the result is a perfect power of 2. Find these three numbers that satisfy this property.

My attempt: I worked out that the three numbers must be either all odd or all even. I tried to expand the formulas based on the numbers being either all odd or all even but was not able to get anywhere and it just got too messy.

I'm guessing there is a trick of some sorts to solve this problem. Any help is appreciated. Thanks!

 

[mfb]

One example should be easy to find. In the easiest case the three integers are all identical, then you have to consider only one case.
I don't know if there are more examples.

 

[Svein]

Let a. b and c be the integers. What you want is values such that $a\cdot b - c=2^{k}$ for some k. The easiest way is to set c=2 and a=2. Then $b=\frac{(2^{k}-2)}{2}=2^{k-1}-1$...

 

[mfb]

Let a. b and c be the integers. What you want is values such that $a\cdot b - c=2^{k}$ for some k. The easiest way is to set c=2 and a=2. Then $b=\frac{(2^{k}-2)}{2}=2^{k-1}-1$...

We also need $c \cdot a - b = 2^m$. With your choice for c and a, that doesn't leave many options, and it leads to the simple example I suggested.

 

[TeethWhitener]

I worked out that the three numbers must be either all odd or all even.

Not necessarily true (consider the case with $2^0$).

I don't know if there are more examples.

You have $a\cdot b = 2^k+c$, where $c$ and $k$ are arbitrary non-negative integers. This puts a simply stated constraint on $a$ and $b$. (Edit: actually nevermind...there's an integer solution for literally any $c$...) (Second edit: I misread the question. Disregard what I said.)

 

[willem2]

Question: There are three non-negative integers with the following property: If you multiply any two of the numbers and subtract the third number, the result is a perfect power of 2. Find these three numbers that satisfy this property.

If 1 is not a perfect power of 2, there are exactly 2 possibilities with a>=b>=c and a<10⁹
If you count 1 as a power of 2 there are 4 examples. All the examples are very small numbers, and will be immediately found with any kind of computer search.
if you know only a and k in $ab - c = 2^k$ there will be only 1 possibility for b and c. (use a>=b>=c)

 

[arpitm08]

So far I've found four solutions:

• 2, 2, 2
• 2, 2, 3
• 3, 5, 7
• 2, 6, 11

I just used Excel to find these. I'm not sure if there are solutions with larger numbers as that would require too many calculations that my laptop couldn't handle.

 

[mfb]

@willem2 excluded further solutions smaller than 1 billion. While that is not a proof, it makes additional solutions extremely unlikely.

 

[willem2]

The problem:
ab-c = 2^k
ac-b = 2^l
bc-a = ^m
If a,b,c is a solution, all permutations are also solutions, so we only have to look for solutions with a>=b>=c. This implies k>=l>=m
What I did, is that once you choose a and k, than since ab-c = 2^k, a(b-1) = ab -a will be smaller than 2^k, this implies that ab is the smallest multiple of a, that is bigger than 2^k, and that means $b = \lceil \frac {2^k} {a} \rceil$.
(b = ceiling (2^k/a) in excel)
c can than simply be computed from ab-c = 2^k and you can than check if the other two numbers are a power of 2.
you can check very quickly if a number is a power of two: n only is a power of two if (n != 0 and (n bitand(n-1)) == 0)

def ispow2(n):
    if (n==0): return False;
    return (n & (n-1)) == 0;

you can let a vary from 1 to max, and check all k with. a <= 2^k < a².