Diophantine Equations | SMO 2024 | Junior | Round 2 |Cheenta | Deepan Dutta

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  • Опубліковано 17 гру 2024

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  • @MatheExtrem
    @MatheExtrem 11 днів тому +1

    7:54 If 5 would divide both u+v and u-v, it would also divide their difference, namely 2v, and therefore v. Since 5 divides both u+v and v, it divides their difference, u. Therefore, let a = u/5, b = v/5. Either 5 divides b or 5 divides a. Therefore neither (a-b) nor (a+b) is congruent to 0 modulo 5. Therefore (a+b)*(a-b) = (u+v)(u-v)/5^2 = 2000/25 = 80 cannot be divisible by 5. A contradiction. Therefore 5 cubed must divide either u+b or u-v.

  • @abhinavkumarverma2224
    @abhinavkumarverma2224 7 днів тому

    We can see that in both (u+v) and (u-v), both are divisible by 5. So ther sum also must be divisible by 5 which is 2v and v is x²+10x+2. We clearly see that 10x is divisible by 5 but x²+2 is not. We can say that x²+2 is not divisible by 5 because the last digit of multiples of 5 are 0 or 5. Therefore x² last digit must be 3 or 8 which cannot be possible.
    Therefore condition not satisfy.

  • @LifeIsBeautiful-ki9ky
    @LifeIsBeautiful-ki9ky 11 днів тому +1

    this method works for specific problems only. It doesnt work for : Y^2=x^4+1, y^2=x^4+x^3+6

    • @Deepan_Dutta
      @Deepan_Dutta 11 днів тому

      That's correct. Actually there isn't any such general method to solve a diophantine equation, for this equation it's one of the strategies and may be for some other equations as well.