A nice and quick elementary number theory problem.

Using elementary techniques, we solve a quick equation.

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[rogerlie]

My solution: It's obvious that a must be a multiple of 3 so set a = 3n, and we get
n + 2b = 3nb
after dividing by 3. Some simple algebra give us
2b/n = 3b - 1
LHS ≤ 2b while RHS ≥ 2b, thus the only solution is when we have equality which means b = n = 1 => a = 3, b = 1.

[thehockyebox]

My solution:
a + 6b = 3ab
3ab - a - 6b + 2 = 2
(a-2)(3b-1) = 2
case 1: a-2 = 1, 3b-1 = 2 => a=3, b=1
case 2: a-2 = 2, 3b-1 = 1 => a=4, b = 2/3
since we want a, b € N, the only solution is a=3, b=1

[goodplacetostop]

9:39

Error in the thumbnail though? a-6b instead of a+6b

[toast_recon]

I appreciate you recognizing the different definitions for natural numbers after this was pointed out some videos back! You have an obvious dedication to clarity.

[deadfish]

Don't need to mess about with gcd if you rearrange it to (a-2)(3b-1)=2. 3b-1>=2, a-2>=-1, so the only solution is 3b-1=2 and a-2=1.

[f-th]

At 7:40, there's quicker way: c = 3b(c-1) means c-1 divides c, but they're coprime, so c-1 = 1, so c = 2. 3 has to divide c, but that's impossible.

[OllieClarke]

Nice puzzle. Another way to do it is to rewrite a + 6b = 3ab as (a - 2)(3b - 1) = 2. So a - 2 is either 1, -1, 2 or -2. Checking those gives you all the solutions: (a, b) = (0, 0) and (3, 1).

[PaulMurrayCanberra]

Surely, pointing out that the gcd of n and n+2 cannot be greater than 2 is a matter of pointing out that multiples of 3 are 3 apart, etc.

[minusonetwelth]

5:00 Doesn't it have to be 3b|a, since a is the product of the two smaller terms 3b and (a-2)?

[AmberZak]

Hi. I think it would be good to clarify that gcd means greatest common divisor. In the
UK we call it highest common factor HCF so it threw me for a bit what gcd stood for. I had to google it.

[issacodegard]

From a = 3b(a-2), we have that a-2 divides a, but that means a-2 is at most half of a i.e. a is at most 4.

[tilek]

3 divides a clearly. a=3k. k+2b=3kb and k=(3k-2)b. Hence 3k-2 less then or equal to k. Hence k=1 and a=3. So b=1 which gives (a, b)=(3, 1)

[noway]

One could also "complete the product" as you have dubbed it before, and garner the same result using the fundamental theorem of arithmetic (unique prime factorization).

[gngeannakakes]

Could you start a couple of series on 1) the Calculus of Variations & 2) Integral Equations?

[zygoloid]

a = 3b(a - 2) implies a-2 | a and 3 | a. From Euclid's algorithm we know that any common divisor of n and n+m divides m, so is ≤ m (if m > 0), so (taking n = a-2, m = 2), a-2 ≤ 2, so 1 ≤ a ≤ 4, so a = 3, b = 1.

[abderrahmanyousfi]

When you said a must divide RHS so a|3b you can also say 3b must divide LHS so 3b|a thus a=3b and replacing this in the equation to solve a then solve b
.

[Ranoiaetep]

My solution:
a = 3ab - 6b
a = 3b * (a-2)
a/(a-2) = 3b
since b >= 1:
a/(a-2) >= 3
a >= 3a - 6
0 >= 2a - 6
a <= 3,
and now you can just try a = 1, 2, 3 and get the answer

[robertgerbicz]

Much easier way: (a-2)*(3*b-1)=2, then abs(3*b-1)<=2 (since a, b are integers), so we can have only b=1, then a=3, which is the only solution.

[leefisher]

Hey, no fair! In the thumbnail it said a - 6b = 3ab, and you're solving a + 6b = 3ab.
In solving the thumbnail, I did pretty much what Adam Romanov did, but with the different sign.
3ab + 6b - a = 0
(3b - 1)(a + 2) + 2 = 0
(1 - 3b)(a +2) = 2
I then drew a graph of a against b to find all solutions (within the scope of the graph)

[crazyAngol]

rearranging you get a=3ab-6b, therefore a must be divisible by 3 so substituting 3k for a you get 3k= 9kb-6b and b=k/(3k-2) if b is a natural number |k|<=|3k-2| and the only natural number that satisfies that condition is 1, therefore a must be 3 and b=1