Thank you for this great episode. This problem is really awesome! And I tell you why. Being also interested in music theory and tuning systems, I found a strong relation to this domain.
Translated into the language of music, this equation asks: How can I divide an octave (which is the frequency ratio of 2 to 1) into three simple intervals where the nominator is one more than the denominator. In just intonation, such intervals are: octave = 2/1, fifth = 3/2, fourth = 4/3, major third = 5/4, minor third = 6/5, major second = 9/8 or 10/9, minor second = 16/15 or 17/16.
For example, the solution a=4, b=5 and c=3 leads to the factors 5/4, 6/5 and 4/3, which are a major third, a minor third and a fourth. If I start at C3, the following notes are E3, G3 and C4. This is a major chord. Great.
Another example, the solution a=2, b=4 and c=15 maps to the intervals of a fifth, a major third and a minor second. Starting at C3 the following notes are G3, B3 and C4.
For me, there is a heavenly connection between mathematics and music.

"I don't want to write it", proceed to write out more than previously needed.

I like very much the way you teach. You experiment with possibilities, instead of being smarty and immediately  presenting the solution. In this way, the student feels like participating in the solution instead of passively watching it. You're a great tracher.

The most charismatic teacher that I have ever seen.

(1+1/3)(1+1/4)(1+1/5) is just beautiful

Awesome reasoning towards getting these solutions.Very much captivating... please continue discussion of such problems.

Never stop learning. I'm 34, out of school since a while, and I loved this video.
Thank you ❤

One of the finest math explanation I've ever seen on Youtube. Excelent job. Man you are a real Teacher.

Great analysis and solution. I have to admit that I viewed this video three time to thoroughly understand your solution. Thank you!!

The problem becomes much easier when you check that 1.25 ^ 3 < 2, which means one of the numbers must be 2 or 3.

Now I want to see the calculus version 😁

Ok people, let’s start over by replacing = 2 by = 3 in the initial proposition.

That was a fun one. I'm trying to take your advice and to never stop learning so that I don't die.

Your handwriting is so good!

Other solution: (a+1)(b+1)(c+1)=2abc. For decomposition, it needs to be for example {(a+1)=b,(b+1)=c,(c+1)=2a} this set of equations leads to a=3 b=4 c=5. There's some other way to factorize, which gives negative or non integer results. The other factorization which gives result is a=b and {(a+1)^2=c,c+1=2a^2}, which leads to a=b=3,c=8.

These questions truly invoke the though process and helps to gain from what is taught in the classroom!! Amazing work by sir!! Thanks a lot. A math question a day keeps boredom away:yougotthis:

Im sure you have been told, but your voice, rythm and handwriting make for a highly enjoyable and almost relaxing video :)
Thank you. Subed

Convert to (a+1)(b+1)(c+1) = 2abc.  ∴ at least one factor on left must be even, say a+1, so a must be odd. Start with a=3 so (3+1)(b+1)(c+1) = 2(3)bc, cancel 2 from both sides to get 2(b+1)(c+1) = 3bc.  Same reasoning as before shows at least one of b or must be even, say b.  Now start at b =2,4…paired with a=3, to see if an integer c can b found. First try a=3, b=2 fails, but next pair (3,4) works and gives c=5. Could go on and try other pairs of a=3, b= even from there,etc.

I really enjoyed the organized and clear explanation of this problem! Broadening the solution space to all integers, notice that two or three of a, b, and c cannot be negative. Otherwise, if all three of them were negative, we'd have a product of three zero or positive fractions each less than 1, which could not multiply to 2, and if two of them were negative, 1 + 1/a would have to equal or be greater than 8, also impossible. Nevertheless, without loss of generality, we could modify the assumption of a being smallest to a being less than or equal to the other positive integer b, retaining 1 + 1/a as the largest factor in the product, leading to the same bound on a. Then, with the same technique, not including permutations, we can find three new solutions: (a, b, c) = (1, 2, -3), (2, 2, -9), and (1, 3, -4).

Love these type of questions and how you explain them