Solving a Diophantine Equation Using an Algebraic Trick 

I said "Accidentally adapted" in the description above because I modified the problem slightly without knowing! 😂😂😂 The original problem in the book is 1/m+1/n+3/mn=1/4. Try solving this and comment down below ⬇ This is a nice Diophantine Equation that can be solved with a little bit of effort. You don't have to use the method that I used in the video. I posted an alternative solution HERE: ru-vid.comUgyIKmrIkUdWgC5CYR54AaABCQ

I’ve heard the technique referred to as “completing the product”

Really cool. I got to the point mn-4m-4n=4 in the video and was playing around with factoring this equation. Factoring it to m(n-4)-4n=4 is where I got stuck, I just couldn't quite get to (m-4)(n-4)=20, that's really the key to seeing what m and n values easily lead to a solution. Still, really great challenge, I really enjoyed it, and I hope that this kind of factoring will help me in future problems (I did save my notes for future reference).

Simply solve for m. m = 4*(n+1)/(n-4) = 4 + 20/(n-4) -> n-4 must be factor of 20.

very cool video! I like the speed with which you explain! (I'm a physicist for reference)

Very good. I tried Vieta's formulas here and it turned out to be a mess, but I did figure out that m+n≥17, so at least I wasn't (completely) wrong.

The old great Simons favorite factoring trick

Finaly i solved it !! Now i will watch your video to know if im right

I had to check the answers for myself on my calculator-so cool!!

There's also the case where both m and n are lesser than 4, such as n=-16 and m=3, which will give you a result of 20

4(m+n+1) = mn Here (m -4)(n-4) = 20 = 5*4 20 = 1*20 implies m = 5, n =24 = 2*10 implies m = 6, n =14 = 4*5 implies m = 8 , n =9 So ( 5,24) (24,5), ( 6,14), ( 14,6) , (8,9), ( 9,8) are the possible solutions

1/m+1/n+1/mn=1/4 Since n and m are not equal to 0, we can multiply everything by 4mn: 4(n+m+1)=mn. (1) (m-4)(n-4)=mn-4n-4m+16 Then (1) (m-4)(n-4)=20 20=1x20=2x10=4x5 So we have the couples of solutions: (5,24), (24,5), (6,14), (14,6), (8,9), (9,8).

another solution: take m < n < mn then 1/m > =1/n >= 1/(mn) , then 1/m + 1/n + 1/(mn)=1/4

In given equation, i just added both sides by 1. It became [1+(1/m)] * [1+(1/n)] = 5/4 & got same solutions by creating fractions where numerator = 1 + denominator

mn-4m-4n-4=0 mn-4m-4n+16=20 (m-4)(n-4)=20 m=5, n=24 m=3, n=-16 m=6, n=14 m=2, n=-6 m=8, n=9 and the permutations of these m=-1, n=0 we have to discard this solution because the original equation will be undefined

make denom mn; m, n cannot equal zero; both have to be pos int 4n + 4m + 4 = mn; 4m-mn=-4n-4 m(4-n)= -4(n+1) m=-4(n+1)/(4-n); n cannot = 4 no pos int for n creates a pos int for m pos int for n make m neg which is not allowed. so if both have to be positive integers; no solutions.

I appreciate your selection of Diophantine equations. When I was in eighth grade, I got my first graphing calculator and started experimenting with parametric and polar graphs. I was selected for a bonus learning experience wherein I would learn on weekends from university professors. I first took a math course, and we were taught about Diophantine equations. However, he never explained them very well, and it was to the point to where I was confusing the term “Diophantine” with “parametric.” That is because we limited ourselves to linear Diophantine equations. The standard procedure for giving solutions to linear Diophantine equations is to find a set of parametric equations for the line with integer coefficients, so an integer parameter maos to an ordered pair of integers. I caught on to the concept so much that I even asked him why he was using “u” for an arbitrary input when I knew that the standard letter was “t”!!!

Yes! If negative solutions also allowed, then add four more.... (m, n) = (3, -16), (2, -6), (-16, 3), (-6, 2) [Actually the third case yields m = 0, which is undefined in the original equation] 😇😇

compi finds 10 integer solutions including negative valued ones: *FindInstance[{1/m + 1/n + 1/(m n) == 1/4}, {m, n}, Integers, 11]*

Hocam 1/m +1/n+1/m^2=1/4 nasil ulaştık.

Nice! I tried substituting 1/m = u & 1/n = v initially and then use SFFT but then realised that that leads to too many fractional possibilities. Hence, your multiplication of 4mn on both sides is the most effective

Don't forget (4,∞) and (∞,4). LOL

I finished the video awesome approach for factoring at the end so many goos tricks in this channel

Pueden considerar n= m+1, tendrán 1/m+1/(m+1)+1/(m(m+1))=4 desarrollan la fracción y obtienen m^2-7m-8=0, factorizan (m-8)(m+1)=0, las soluciones son m=8 y m=-1, como se pide m>0, la soluciones son m=8, n=8+1=9

My correct but flawed factoring of (m+1/m)(n+1/n) = 5/4 dîd guide me in what values to check and I found most of the solutions quite quickly. I could tell when the numbers to check had got too large.

No way I could have done that, I have no coloured pencils.

Another way is to solve for m and we get m=(4n+4)/(n-4)=4+20/(n-4), so n-4 must be a divisor of 20. And we get the same solution.

I solved it by CND

I also have the same answer

M=9 n=8 or n=9 m=8

f(x-y) = f(x) - f(y) is equivalent to the definition of a linear function, so no numerical substitutions necessary. Just assume f(x)=mx immediately getting m=5/2 which leads to f(16)=40. The linear ones are straightforward. Give us some non linear functional equations :) Thanks for the problem!

Good one

8 and 9 Answer one has to be smaller than 1/8 since + 1/8 = 1/4.. so 1/8 + 1/9 + 1/72= 1/4. So 8 and 9 (or 9 or 8) is just one of a number of solution?

Very interesting.

Simon's factoring trick is always useful : )

1:43 I don't get it why can't m=n

I love integer solution good choice !!!!!😍

Good problem, took me 10 minutes 👍

saying "simon's favourite factoring trick" is very tiring; how about we shorten it to SFFT ?

Do gaussian integers count?

Why you always want that we comment and subscribe your videos?

good solution!!