Solving an Interesting Diophantine Equation in Number Theory - Math Olympiad

In this video, I am presenting step-by-step solution to this interesting diophantine equation. After a few steps, we come up with a surprisingly easy solution to this diophantine equation. This video contains very important skills to perform when solving a diophantine equation, and it will help you solve more complicated diophantine equation. This question will also be a good problem for math olympiad if you prepare one. Come check this question and watch it until the end. You will get a skill to be used for almost any type of diophantine equation. More to come! Stay tuned!

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[zsoltnagy]

I'm more fan of the *Lagrange Method:*
Maximise the function f(x, y)=x+y under the condition g(x, y)=1/x+1/y-2/15=0.
So the Lagrangian L is the following:
L(x, y, λ)=f(x, y)-λg(x, y)=x+y-λ(1/x+1/y-2/15)
with the Lagrange multiplier λ.
f has the same critical points as the Lagrangian L at (x|y) (and some λ).
So
I: ∂xL=1+λ/x²=0
II: ∂yL=1+λ/y²=0
III: ∂λL=1/x+1/y-2/15=0
From III it follows, that 1/y=2/15-1/x or 1/y²=(2/15-1/x)²
and from I it follows, that λ=-x².
So we get the following with II:
*1+(-x²)(2/15-1/x)²=0*
⇔ 1-x²(4/225-4/(15x)+1/x²)=0
⇒ x²-15x=0
⇔ x(x-15)=0
SInce of course x≠0, therefore we only get x=15.
From III it follows with x=15, that y=15. (Also from I it follows with x=15, that λ=-225.)
*So f(x, y)=x+y is maximal at (15|15) with fmax=f(15|15)=15+15=30.*
Sure, x and y were originally supposed to be integers and for the Lagrangian L they became or further assumed to be reel continuous numbers.
But since integers are a subset of real numbers and since this method entails the solutions for x and y to be actually integers, therefore this is also a legitimate method to get the answer to the proposed question.

[lachymcgarry]

This may have been mentioned, but this question is very easy if you just use derivitives. If you rearange, you get : y= (15x)/(2x-15). Then if you let some number A= x + y, you can substitute y in to get: A = (2x^2)/ (2x-15). Thus, dA/dx = (4x^2 - 60x)/(2x-15)^2. If you let dA/dx =0 you get x= 0 and x= 15 as answers ( obviously 0 doesnt work so u can ignore it). Then sub x= 15 back into A and you get 30.

[Min-cvnt]

Love diophantine equation. Love your videos

[MrGLA-zsxt]

Truly love your video. I worked with quadratic equation too

[crustyoldfart]

In my simple clod-hopper approach, the expression ( x + y) / ( x * y ) would suggest that x and y are interchangeable, which in turn implies x = y, so we can write : ( x + y) / ( x * y ) as 2*x. / x^2 -> 2 / x = 2 / 15 -> x = 15 = y. This may be simplistic but it would seem that it leads to a correct result, albeit without rigorous proof. And as to that I have no intention of spending time finding such a proof.

[flybyw]

By simply looking at 1/+1/=2/, it is obvious x=y=15; can't get any integer smaller.

Since 2/((x+y)/xy)=2/(2/15)= 2xy/(x+y) = 15 is the harmonic mean of x and y the smallest sum x+y to achieve this is when x+y=2•15 and x=y. So x+y=15+15=30. {For example 1/30 + 1/10 = 1/30 + 3/30 = 4/30 = 2/15 because 2•30•10/(30+10) = 600/40 = 15, but 30+10>15+15 so that's no good.}.

[jagmarz]

On the face of it, if x = y = 15, we have x+y = 30 and we otherwise satisfy the equation, so we're looking for some other x+y<30. OK, so we clear the denominators on the left to get x+y = 2xy/15, which is an integer. So xy must have a factor of 3 and a factor of 5. If both the 3 and 5 are in x or y, then we have x=y=15, which we already know. So (wlog), lets arbitrarily assign x=3a and y=5b to account for the factors we need. However, 2/15 < 1/5 < 1/3, so we cannot have x or y be 5 or 3, and thus neither a nor b can be 1. Now b=3 corresponds to x=y=15, b=5 would put x(=3*(minimum a = 2)=6) + y(=5*5=25) =31 > 30, and b>5 would be even larger. So we only have to check b=2 or b=4. If b=2, then y=10 and 1/x = 2/15 - 1/10 = 1/30. But then x+y = 40 (>30, so not good). Similarly, if b=4, then y=20, and 1/x = 2/15-1/20 = 5/60 = 1/12, but that still leaves x+y=32 (>30), and the minimum is the obvious x=y=15, so x+y = 30.

[FatihKarakurt]

Easier with AMGM.
(x+y)/2 >= sqrt(xy). At minimum should be equal to sqrt(xy). Which is also xy/15. Therefore sqrt(xy)=15. So (x+y)=30.

[miro.s]

There is much straigtforward and much faster solution using Vieta's formula

[Nerthos]

They're just 15s from different races