Solve for all positive integers [Diophantine]

[@ricardoguzman5014]

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.

[@FirasFaleh-oj2ex]

Nice

[@peshepard412]

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

[@ΠαναγιώτηςΣιφωνιός]

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

[@RobertChan2003]

Solve y=x^5 and y=3x^2+216 (parabolic formula y=ax^2+c) by graphic method, there can only be one intersection point and also can be visualized either 3 or 4 (told to find positive integer). By substitution, when x=3, y=243 for both equations.

[@pijanV2]

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

[@randomjin9392]

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.

[@MadaraUchihaSecondRikudo]

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

[@abhizith_1]

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

[@GreenMeansGOF]

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

[@PrimeNewtons]

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

[@chaosredefined3834]

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

[@gregoryknapen9133]

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]

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]

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]

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]

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]

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

[@emilie375]

Z+ ? What happened to IN ?

[@narangfamily7668]

Very nice. Loved it…..

[@flexeos]

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]

Rational root theorem would get us there more mechanically

[@chuckc3665]

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

[@adgf1x]

x=3ans

[@Maths786]

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 😢

[@antonionavarro1000]

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

[@Better-v8q]

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]

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]

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

[@PhilLeChatounet]

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]

Indians assemble here❤

[@pclown9490]

Dude just use synthetic division😂😂

[@Maths786]

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]

Tonight