A Nice Diophantine Equation | Putnam 2018 A1

#MathOlympiad #Number Theory #Putnam
Here is the solution to Problem A1 in the Putnam Mathematical Competition 2018!
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
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[@franciscogallegosalido1402]

I think you can reduce the amount of work by simply asuming that a < b (a cannot be equal to b) without losing any generality.

[@ceooflslam]

Nice elementary problem, good solution also 👍

[@letsthinkcritically]

Thank you!!

[@satyapalsingh4429]

Heart filled with joy . Very nice method of finding the integral values of a and b . Thanks

[@tonystarklive9018]

Can you solve a beautiful combinatoric problem that you know ,I like combinatoric

[@letsthinkcritically]

Sure!!

[@mrityunjaykumar4202]

it can also be done by considering two cases, 1: (a,b) are co prime, and solving it... it turns out to be impossible.. 2: not co-prime => a=dA, b=dB where (A,B) are co-prime. then putting it in the eqn...we get A=1, B=2, d=1009..since d(A+B)/(d^2(AB))=(A+B)/dAB.. now A+B=3.. only if A=1 and B=2 and vice versa.. hence a=dx1, b=dx2 => 3/(dx2)=3/2018 => d=1009 => a=1009, b=2018 so odered pairs =(1009,2018), (2018,1009)

[@BabasCrafter]

Great Video. One more question. I've heard that we can get only 3 (or in form (b,a) 6 solutions.) using the form 1/a + 1/b = 3/n . Have you an idea why this is the case?

[@RexxSchneider]

If you follow the method used in the video, you will arrive at this: (3a - n)(3b - n) = n^2 That results in (3a - n) being any of the possible factors of n^2 (and 3b-n being the complementary factors). If n is a prime > 3, then there are just three factors: 1, n and n^2, so there are at most three possible solutions: a = (n+1)/3; a = 2n/3; a = n(n+1)/3. However, not all of those are possible because the first one and third one require n ≡ 2 mod 3 to make a an integer, while the second is not possible because n is prime. So you will have either two (or four counting (b,a) as well as (a,b)) solutions for prime n ≡ 2 mod 3 or no solutions for prime p ≡ 1 mod 3. The scenario in the video has n equal to 2p, where p is a prime > 3. That yields nine possible factors of n^2 (since n^2 is equal to 4p^2): (3a - n) = { 1, 2, 4, p, 2p, 4p, p^2, 2p^2, 4p^2 } If p ≡ 1 mod 3, then n ≡ 2 mod 3 or -n ≡ 1 mod 3, so the possible factors become { 1, 4, p, 4p, p^2, 4p^2 }. So a = { (n+1)/3, (n+4)/3, n/2, n, n(n+4)/12, n(n+1)/3 }. If p ≡ 2 mod 3, then n ≡ 1 mod 3 or -n ≡ 2 mod 3, so the possible factors become { 2, p, 4p, 2p^2 }. So a = { (n+2)/3, n/2, n, n(n+2)/6 }. I think that shows that what you heard isn't true.

[@RAJSINGH-of9iy]

Which setup do you use???please let me know about that.

[@TechToppers]

Can you please tell me how you make Thumbnail by Latex?

[@letsthinkcritically]

I type in LaTeX and capscreen so to form images

[@chowy7468]

What if 3a-2018 and 3b-2018 are both negative?

[@cosimodamianotavoletti3513]

Noticing that both factors have to be congruent to 1 mod 3 we have only two cases, namely 3a-2018=3b-2018=-2018, that is a=b=0 (rejected), or 3a-2018=-2*1009^2 and 3b-2018=2, which yields a solution with a

[@chowy7468]

@@cosimodamianotavoletti3513 thanks

[@quirtt]

Op 👍

[@ahmedabdelwahab9085]

I didn't understand anything after 0:00

[@ricardocesardasilvagomes9549]

UAU....

[@mctrafik]

WHY the congruence to 1 mod 3???!?!?!?!?!? No video about this problem seems to explain this. WTF? Downvote.

[@ericzhu6620]

for example, since a is integer, 3a is congruent to 0 mod 3, and 2018 is congruent to 2 mod 3, so 3a - 2018 a=can be only congruent to 1 mod 3, it's the same with b

[@advaymayank1410]

3a - 2018 has to be divisible by 3, since a is an integer. so if you have 3a-2018 congruent to n modulo 3, if you add 2018 to both sides, you will see that 3a congruent to n + 2018(n + 2018 is divisible by 3) modulo 3. In such a case, 3a - 2018 has to be congruent to 1 example - 3a - 2018 = 1 3a = 2019, 2019 is divisible by 3.

[@raffi.88]

No se te entiende nada.