Solve for all positive integers [Diophantine]

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  • เผยแพร่เมื่อ 24 ธ.ค. 2024

ความคิดเห็น • 34

  • @ricardoguzman5014
    @ricardoguzman5014 วันที่ผ่านมา +11

    Products are ingenious. Here is my solution:
    X⁵ - 3X² - 216=0; add 3X² + 216 to both sides of the equation:
    X⁵ - 3X² - 216 + 3X² + 216 = 3X² + 216
    X⁵ = 3X² + 216; 3 can be factored from the terms on the right side of this equation
    X⁵ = 3(X² + 72); now since X⁵ must be divisible by 3, only multiples of 3 need be tested for X: 3, 6, 9,12,...etc.
    testing 3: 3⁵=3(3² +72)
    243 = 3(9+72)
    243=3(81)
    243=243; 3 checks out
    testing 6: 6⁵=3(6² + 72)
    7,776=3(36+72)
    7,776=3(108)
    7,776=324; 6 fails since 7,776≠324
    No other of the integers need to be tested since it is immediately evident that it's fifth power is going to be much larger than 3 times it's square + 216, because 6 already yielded a number much larger, and anything greater than 6 will yield a number larger still, so 3 is the sole solution.

  • @peshepard412
    @peshepard412 วันที่ผ่านมา +4

    By Descartes law of Signs, there is only 1 positive root or zero.

  • @ΠαναγιώτηςΣιφωνιός
    @ΠαναγιώτηςΣιφωνιός วันที่ผ่านมา +3

    Since the solution is a integer it should be a divisor of 216 by trial and error we get x=3 as solution. Then by Horner method the factors of x^5-3x^2-216 are x-3 and x^4+3x^3+9x^2+24x+72. The latter one has no positive solutions therefore x=3 is the only positive integer solution

  • @GreenMeansGOF
    @GreenMeansGOF วันที่ผ่านมา +11

    The number 1 wasn’t invited to the perfect square party.😔

    • @PrimeNewtons
      @PrimeNewtons  วันที่ผ่านมา +5

      That was obvious prejudice against such an innocent number. I apologize.

    • @chaosredefined3834
      @chaosredefined3834 วันที่ผ่านมา +3

      To be fair, while it was an oversight... Pretty clearly, 1 - 3 - 216 =/= 0.

  • @RobertChan2003
    @RobertChan2003 วันที่ผ่านมา +1

    Plot y=x^5 and y=3x^2+216 (parabolic formula y=ax^2+c), then only one intersection point can exist and that is either 3 or 4 (told to find positive integer). By substitution, when x=3, y=243 for both equations.

  • @antonionavarro1000
    @antonionavarro1000 16 ชั่วโมงที่ผ่านมา +1

    You forgot to check x = 1
    216 = 1³ • 2³ • 3³
    You only need to veify three cases for the factor x³-3:
    case x=1
    x³ - 3 = 1³ - 3 = -2
    No valid vale cause x² (x³-3)=216 and x² is positive and x³-3 is negative.
    Case x=2
    x³ - 3 = 2³ - 3 = 5
    No. Five is not a factor of 216.
    Case x=3
    x³ - 3 = 3³ - 3 = 3 (3² - 1) = 24
    Found. 24 is factor of 216.
    Test: x² (x³ - 3) = 3² • 24 = 216

  • @pijanV2
    @pijanV2 23 ชั่วโมงที่ผ่านมา

    7:02 you can usethe factor theorem also just to see if you get any other solutions

  • @randomjin9392
    @randomjin9392 วันที่ผ่านมา +1

    Reducing this mod 3 we have x⁵ = 0 mod 3, meaning x = 0 mod 3. Set x = 3k for an integer k. We arrive at 9k⁵-k²-8 = 0. The sum of coefficients is 0, thus k = 1 is a solution. It easily factors: 9k⁵-k²-8 = 9k⁵-9k²+8k²-8 = (k-1)(9k⁴+9k³+9k²+8k+8). But 9k⁴+9k³+9k²+8k+8 = (3k²+3k/2)² + (2k+2)² + 11k²/4 + 4 > 0 for any k. So the quartic has no real roots and for k = 1 we get x = 3 as the only solution.

  • @abhizith_1
    @abhizith_1 วันที่ผ่านมา +3

    i am wathing you from india i love your videos a lot

  • @MadaraUchihaSecondRikudo
    @MadaraUchihaSecondRikudo วันที่ผ่านมา +1

    You could have checked the solutions easier by seeing if the x you picked satisfied x^3-3 = , instead of plugging it into the origin equation

  • @gregoryknapen9133
    @gregoryknapen9133 วันที่ผ่านมา

    We know that (x^3 - 3) divides 216=6^3. And x^3 must be smaller than 6^3. So it is just a matter of testing cubes less than 6^3. Working backward from x=5, we quickly find that x=3 is the only solution.

  • @itsphoenixingtime
    @itsphoenixingtime วันที่ผ่านมา

    I should have known the diophantine part was relevant, since I tried to solve it like a normal quintic.
    I guessed x = 3 as one of the roots and then divided and got a polynomial with all positive coefficients. By that I reasoned that if that polynomial has any real roots, they must be negative since all the terms are positive [so they couldn't add up to 0 in any way].
    That reasoning should have been enough for me to know that x = 3 was the only real root.

  • @rahulbamel
    @rahulbamel วันที่ผ่านมา +1

    Sir I am an Indian 14 year old student. thank you for explaining all your videos in very curious and fun way😊😊. I have very keen interest in mathematics. and through your channel I got a medium to study more about maths
    can you make a video on following problem?
    If sin A, sin B, sin C are in AP and cos A, cos B, cos C are in GP, then
    Value of (cos²A+Cos²C - 4 Cos A•CosC )/1 - Sin A•Sin B
    if video is not possible then can you please explain it in comments

  • @emilie375
    @emilie375 วันที่ผ่านมา

    I used bijection theorem (don't know the english name) : I caluculate the derivate and then I saw there is only one real solution, greater than 1.
    Then, I also used the rationnal root theorem : if a positive integer is solution then it must divide 6^3. It works for x=3 □

  • @Kosekans
    @Kosekans 17 ชั่วโมงที่ผ่านมา

    Is it not true, that integer solutions are divisors of the constant term? Would brute force not work? Simply try +/-1, +/-2, +/-3, +/-6 -> done?

  • @maths01n
    @maths01n วันที่ผ่านมา

    Let's get into the video... ❤ arousing my interest ever to watch More and more

  • @emilie375
    @emilie375 วันที่ผ่านมา

    Z+ ? What happened to IN ?

  • @narangfamily7668
    @narangfamily7668 วันที่ผ่านมา

    Very nice. Loved it…..

  • @flexeos
    @flexeos 21 ชั่วโมงที่ผ่านมา

    it is easy to see that he function f(x)=x5-3x2 has a strictly positive derivative for integer x>1. f(2)=20 is too small , f(3)=216 is a solution and so is the only solution as as f'>0 for integer x>1.

  • @MichaelGrantPhD
    @MichaelGrantPhD วันที่ผ่านมา

    Rational root theorem would get us there more mechanically

  • @chuckc3665
    @chuckc3665 วันที่ผ่านมา

    f(2^x )+f(2^(-x) )=1,f(2)=? can you solve this? thanks

  • @Maths786
    @Maths786 วันที่ผ่านมา

    Sir please do a limit question which was came in
    JEE Advanced 2014 shift-1 question number 57
    it's a question of a limit
    lim as x-->1
    [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
    You have to find the greatest value of a
    It has 2 possible answers 0 and 2
    But I want the reason that why should I reject 2 and accept 0
    Because final answer is 0
    Please help 😢

  • @adgf1x
    @adgf1x วันที่ผ่านมา

    x=3ans

  • @pclown9490
    @pclown9490 16 ชั่วโมงที่ผ่านมา +1

    Dude just use synthetic division😂😂

  • @Better-v8q
    @Better-v8q วันที่ผ่านมา +3

    Instead of putting the the number in the question and checking if it statisfies the equation you could've checked by using the factor (x3-3)
    For example,
    For x=2, instead of finding f(2), put x3= 54+3. x≠integer, hence rejected.
    It will make the solution faster.

  • @prollysine
    @prollysine วันที่ผ่านมา

    x^5+/-x^4+/-x^3-3x^2+/-x-216=0 , (x-3)(x^4+3x^3+9x^2+24x+72)=0 , / x^4+3x^3+9x^2+24x+72=0 , complex roots , not a solu / ,
    1 -3 solu , x=3 ,
    3 -9
    9 -27
    24 -72
    72 -216

  • @chrismcgowan3938
    @chrismcgowan3938 วันที่ผ่านมา

    x = 3 works Just start with x =1 and test, then x = 2 then test , x= 3 then test too easy

  • @PhilLeChatounet
    @PhilLeChatounet วันที่ผ่านมา

    x²(x^3 - 3) = 4*54
    x² = 4, x = 2, so x^3 - 3 = 5, not 54
    more easy to see if it's a solution :p

  • @HrishikeshRaj-qv6lr
    @HrishikeshRaj-qv6lr วันที่ผ่านมา +2

    Indians assemble here❤

  • @Maths786
    @Maths786 วันที่ผ่านมา +2

    Sir please do a limit question which was came in
    JEE Advanced 2014 shift-1 question number 57
    it's a question of a limit
    lim as x-->1
    [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
    You have to find the greatest value of a
    It has 2 possible answers 0 and 2
    But I want the reason that why should I reject 2 and accept 0
    Because final answer is 0
    Please help 😢

    • @PrimeNewtons
      @PrimeNewtons  23 ชั่วโมงที่ผ่านมา

      Tonight