A Nice Equation with Integer Solutions

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#NumberTheoryProblems #DiophantineEquations

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[leickrobinson]

I think this approach might be easier (and it gets all three solutions):
On the left, as you noted, we have a sum of three powers of two,
2^a + 2^2b + 2^3c
This motivated me to look at the unique binary representation of 328,
328 = 0b101001000 = 256 + 64 + 8
= 2^8 + 2^6 + 2^3
Since the binary representation of an integer is always unique, then we just need to match up the three powers of two on the left with the three powers of two on the right.
I.e., map {a, 2b, 3c} to {8, 6, 3}.
There are three integer solutions:
(a=8, b=3, c=1)
(a=6, b=4, c=1)
(a=3, b=4, c=2)

[fabiog]

In this case I solve this problem with a table solution : 2^a =2, 4, 8, 16, 32, 64, 128, 256 4^b=4, 16, 64, 256 8^c=8, 64 . The only way to get 328 is by summing 8+64+256 from the table listed above there are 3 possible permutations of 8, 64, 256 :
A=8, b=3, c=1
A=6, b=4, c=1
A=3, b=4, c=2

[zh]

My solution:
1) Reduce problem to
2) Note that z has to be divisible by 3, otherwise c in not an integer
3) Note that y has to be odd, otherwise b in not an integer
3) Suppose all x, y, z>0 then LHS is even while RHS is odd. Contradiction, so at least 1 of x, y, z is 0
4) Suppose 2 of x, y, z is 0 then LHS is even while RHS is odd. Contradiction, so exactly 1 of x, y, z is 0
5) WLOG (By symmetry) suppose z=0 then 2^x+2^y = 41-1 = 40 and assume x<=y
6) Suppose x=y then 2^x+2^y = 2*2^x = 2^(x+1) = 40, no integer solutions, hence x<y
7i) Note that 32 < 40 = 2^x+2^y < 2^y+2^y = 2^(y+1), y+1 > 5, i.e. y >= 5
7ii) If y>6 then 40 = 2^x+2^y > 2^x+64 > 40, which is a contradiction, i.e. y <= 5. Thus y = 5
8) Thus 2^x = 40-2^5 = 8 and x = 3
9i) If (x, y, z)=(3, 5, 0) then possible values of z is 0, 3 (by result of (2))
9i) If (x, y, z)=(3, 5, 0) then possible values of y is 3, 5 (by result of (3))
10) If z = 3 then 3c-3 = 3, i.e. c = 2 and y = 5, which implies b = 4 hence (a, b, c) = (3, 4, 2)
11i) If z = 0 then 3c-3 = 0, i.e. c = 1 and y = 3, which implies b = 3 hence (a, b, c) = (8, 3, 1)
11ii) If z = 0 then c = 1 (from (11i)) and y = 5, which implies b = 4 hence (a, b, c) = (6, 4, 1)
There are 3 solutions: (x, y, z)=(3, 4, 2), (6, 4, 1) and (8, 3, 1)

[vacuumcarexpo]

I think the integer solutions of this equation are (a, b, c)＝(3, 4, 2), (8, 3, 1), (6, 4, 1).

Am I wrong?

[jacobdrum]

Divide both sides of the equation by 8 to get 2^(a-3)+2^(2b-3)+2^(3c-3)=41. The only way to write 41 as a sum of three powers of two is as 1+8+32. These exponents are 0, 3, 5. Thus, (a-3), (2b-3), (3c-3) must be of some permutation of 0, 3, 5.

From here, realize that 2b-3 can’t equal 0, and 3c-3 can’t equal 5. This eliminates 3 possible solutions. The remaining are:
(3, 4, 2)
(8, 3, 1)
(6, 4, 1)

[mathematicskids]

the what about the other possible solution
a=8, b=3, c= 1
and a = 6, b= 4, c= 1
and a=3, b=4, c=2

[snejpu]

I found 3 solutions using brute forcing : ) Since, you know, c might only be 1 or 2, b might only be from 1 to 4 and a from 1 to 8.

[MathElite]

Wow interesting, I like myself some number theory
And 18K subscribers complete !!

[Goku_is_my_idol]

First we check mod3
(-1)^a + 1 + (-1)^c = 1( mod3)
a and c are either even-odd or odd-even
Max. Values each can take is easy to calculation since max. Power of 2 possible which is less than 328 is 256
So a<=8 b<=4 c<=2
First lets take c odd
Then a has to be even
So c=1
Substract 8 from both sides
2^2*a0 + 4^b = 320
4^a0 + 4^b = 320
Let one be bigger and other be smaller
For convenience take k1 and k2 with k1>k2
4^k2(4^(k1-k2)+1)= 64*5
k2= 3
k1= 4
So here itself we get two cases
(a, b, c) = (6, 4, 1), (8, 3, 1)
Next take c=2
a has to be odd
2^a + 2^2b0 = 264
Same method
k1 and k2 with k1>k2
2^k2 (2^(k1-k2)+1)= 2³ * 33
So k2=3
k1=8
So here we get (a, b, c)= (3, 4, 2)
Other case not possible cuz a has to be odd
(Here a=2a0 and b=b0. God knows why i subbed b like an idiot)

[Qermaq]

My approach was kinda backward to yours. I first noticed the only integer powers of 8 that are less than 328 are 8^1 and 8^2, so I only need to find two powers of 2 that total to either 320 or 264, Quick examination shows that 264 can be split into 8+256; 8 is not a power of 4 so we get 2^3 + 4^4 and a=3 b=4 c=2. The other option, 320, can be split into 64+256, and as either could be a power of 4, we get a=6 b=4 c=1 and a=8 b=3 c=1.

[vladimirkobarov]

The answer you showed is not complete

(a, b, c) = (6, 4, 1) and (8, 3, 1) are also solutions

[armacham]

I was very careful, I looked at all possibilities (including negative integer values for a, b, c) and got the three solutions.

Here is where Syber went wrong: he is assuming that each part of this must be an integer:
(1 + 2^(2b-a) + 2^(3c-a))

For example, here is one solution he missed (a=8, b=3, c=1). Subbing that in, we get:

2^8 * (1 + 2^(6-8) + 2^(3-8)) = 256 * (1 + 2^-2 + 2^-5) = 256 * (1 + 0.25 + 0.03125) = 256*1.28125 = 328

He incorrectly assumed that A could not be bigger than 3 because he assumed the contents of the parenthesis must be an integer. But he also incorrectly assumed that A can't be less than 3.

What if a = 2 and the contents of the parenthesis are equal to 2 * 41? That would also be a valid solution. It turns out there's no way to get (1 + 2^(2b-a) + 2^(3c-a)) to be equal to 2*41. But it was incorrect to dismiss that possibility out of hand. Similarly if the contents could be equal to 4*41 then you could have a=1. Or if the contents could equal 8*41, then you could have a solution with a=0.

[holyshit]

I wrote C# program and got
a= 3, b= 4, c= 2
a= 6, b= 4, c= 1
a= 8, b= 3, c= 1

[lavneetjanagal]

A cleaner way to solve this is to write everything in base 2. 328 = 101001000 and the expression on LHS is 2^a+ 2^(2b)+2^(3c), i.e., in base 2 the number has '1' at a+1, 2b+1, 3c+1. Looking at the base 2 representation of 328 we see that a, 2b and 3c can only take values from 3, 6 and 8 (because the 1 is at 4th, 7th and 9th place respectively). 'a' can take any value but 3c can only be 3 or 6 giving c=1 or 2. When 3c=3, b has 2 possible values because other two numbers are even, i.e. either b=3 giving a= 8 or b=4 and a=6. On the other hand when 3c=6, b has only one possible value because only one of the other numbers is even, i.e., b=4 and a=3. So the full solution set is
a=3, b=4, c=2
a=6, b=4, c=1
a=8, b=3, c=1

[henrybarber]

Can’t you write 328 as the sum of 3 powers of 2 (which are 2³, 2⁶ and 2⁸) and exhaust all the possibilities that way? That gives you the other two sets of solutions: a = 8, b = 3, c = 1 and a = 6, b = 4, c = 1

[travisgreen]

Software engineer here, so I am usually thinking in binary 🙂

In binary, a power of 2 is has exactly 1 set bit. Raising a power of 2 to an integer power has the effect of shifting all the bits in the same direction (to the left for positive and the to right for negative); the quantify of set bits does not change, e.g. 8 = 1000b and 8^2 = 64 = As 328 = 101001000b, there are exactly 3 bits set. Adding three powers of 2 will produce a binary number with three bits set, unless two or more of the powers of 2 the same for which the quantify of sets bits can be one or two, e.g. 10b + 10b + 1000b == 1100b and 10b + 10b + 100b = 1000b. Since 328 has three bts set and there are three power of 2 addends, the power of 2 addends must be distinct. So, the problem can be restated as finding integers a, b, and c, where each of {a, 2b, 3c} are elements of {8, 6, 3} and a != b != c. There are 6 combinaions of the integer powers of the unique binary factorizatoin of 328: {8, 6, 3}}. 8 is not divisible by 3, so 3c cannot equal 8 is c is an integer, that eliminates 2 of the 6 combinations. 3 is not divisible by 2, so 2b cannot equals 3 if b is an integer; this would eliminate 2 of the 6 combinations, but one of the two was already exlucded from the 8 not being divisible by 3, so just 1 more is eliminated. Therefore, there are 3 (6 - 2 - 1) remaining possibible cominations, for which the (a, b, c) = { (3, 4, 2), (8, 3, 1), (6, 4, 1) } are shown to work.

[drpkmath]

Great job my friend! Number theory is one of my favorite topics

[-basicmaths]

We can also write
a=2b+m, 3c=2b+n
a=3c+m, 2b=3c+n.And find other 2solutions.

[Jeeadv-bvlm]

Wow.Its been a month im watching your videos...my math skills are on another level now💪🏼Thank You!

[iwd]

My solution was a=8, b=3 and c=1. I searched for the highest possible a. 2^8 makes 256 and the rest of 72 can be written as 64 and 8. Both fit the bases of 4 and 8.