A Beautiful Equation From Samara Math Olympiads

[@Flagellatore9]

X = 1 is a solution. it’s The only One because every number > 1 gives a number too big. If x

[@SyberMath]

Isn’t this a really cool problem? 😀

[@Flagellatore9]

@@SyberMath “Always” (Read it with The sad voice of Severus Snape)

[@alexanderbrinkley4332]

@@SyberMath no

[@Flagellatore9]

@@alexanderbrinkley4332 why not?

[@easymathematik]

Isn‘t the solution obvious? x=1. Its the only solution, because x^odd = mx+ b has exactly one solution, if m

[@MattyFein04]

Exactly... should take no more than 15 seconds to figure this out

[@broytingaravsol]

or x^19^95 always goes much and much higher than 1996x if x>1

[@chinmay1958]

let y = the 19^95th degree polynomial in x. we see that y = 0 only if x = 1 because y'(x) is always positive. What this means is that y is monotonically increasing and hence cuts the x axis only once.

[@surendrakverma555]

Good. Thanks 🙏

[@gregwochlik9233]

Looking at the thumbnail, for about 5 seconds, x = 1. x's exponent is really huge, but 1^(anything *) is 1. 1996 + 1 = 1997. 1996 was the year of my matric (final year of high school). 1997 was my first year of my engineering studies.

[@freddyalvaradamaranon304]

Muy interesante video de una singular ecuacion ,muchas gracias por compartir el procedimiento 😊❤😊.

[@scottleung9587]

Cool!

[@sabrenhatam4145]

شكرا جزيلا عله كل امثاله وان اشاء الله يعطيك الصحه والعافيه والموفقيه

[@jacobs31milkacapuccino]

Eqastions as eq. in video in video aren't too hard, we just must to find some string, which can show the solution for us

[@Psykolord1989]

Before watching: Let's check the special cases first; 0 and ±1. 0 does not work (0^n = 0, so we would have 0+0 = 1997) -1 does not work (-1-1996 is not 1997...but it IS -1997, which gives us hope for the third case... 1 gives us 1^(19^95) + 1996(1) = 1 + 1996 = 7. Thus, 1 is a solution. Now for the "regular" cases. Negative solutions do not exist. We would have the sum of two negatives equaling a positive. Numbers between 0 and 1 are not solutions; the x^(19^95) term would always be less than one, and 1996x would always be less than 1996, and the sum of something less than 1 and something less than 1996 is going to be less than the sum of 1 and 1996; therefore, any X between 0 and 1 gives a result less than 1997 Numbers above 1 are not solutions; you would have a sum of something larger than 1 and something larger than 1996, which is larger than the sum of 1 and 1996, which is 1997; therefore, any X above 1 gives a result greater than 1997. With the "regular" cases providing us with no solutions, our only solution is X=1

[@KRYPTOS_K5]

Excellent!!! Brasil (ps what is Samara Olympiada and where is it in math literatura?)

[@SyberMath]

Thank you! It's in Russia

[@KRYPTOS_K5]

@@SyberMath Thank you. I know if they have a public archive of tests? Congrats for your channel

[@devondevon4366]

1 x^ 19^95 + 1996x = 1997 x^ 19^95 - 1 = 1996 - 1996x = 1996 ( 1-x) when x= 1 1 - 1 = 1996 )(1-1) 0 = 0 x= 1

[@AltAaltonnov]

Shouldn't it have been stated as: find all real roots (or something like that))...

[@vybs9235]

For x>1, it tends to infinity And for less than 1 it tends to 0 hence only condition left, is that it must he equal to 1. Btw can u take a few questions from the RMO? Indian regional mathematical olympiad

[@ИванЛебедев-л3н]

Задача лëгкая, хотя я перенëс 1996х в правую часть, после чего даже анализа производных особо не понадобилось

[@nasrullahhusnan2289]

For simple writing let a=19⁹⁵. x^a+1996x=1997 mod(x^a,1996)=mod(1997,1996) [mod(x,1996)]^a=1 mod(x,1996)=1 --> x=1996k+1 where k is an integer. For k=1, x=1. As check: 1^a+1996(1)=1+1996 =1997 For greater value of k, x becomes large and x^a is extra-ordinarily large numbef

[@proffessorclueless]

You haven't checked the fractions.

[@nasrullahhusnan2289]

@@proffessorclueless: What fraction?

[@proffessorclueless]

@@nasrullahhusnan2289 All of them. The video shows why there aren't any.

[@-basicmaths862]

19^95=t, is a odd number. x^t+1996x=1997 x^t-1+1996x-1996=0 (x-1)z+1996(x-1)=0 (x-1)(z+1996)=0 x-1=0, x=1

[@rakenzarnsworld2]

x = 1

[@fahrenheit2101]

So many guess and check equations... x=1 is the obvious solution, and verifying very easily that we're dealing with an increasing function guarantees a unique root. In the complexes, I wouldn't even consider daring to approach this.

[@MariaJohnson-v7b]

Yundt Parkways

[@hmkl6813]

Now find all solutions😈

[@SyberMath]

😜😁

[@raghvendrasingh1289]

Let 19^95 = n (0dd positive integer) Eq will be x^n+1996x-1997= 0 Obviously x=0 is not a solution. Here f(x) = x^n+1996x -1997 ( one change in sign of coefficients) and f(-x) = -x^n -1996x-1997 ( no change in sign) Hence by Descartes' rule of signs there is at most one positive root and no negative root. 1+1996-1997=0 hence x= 1 is the only real root which is positive

[@proffessorclueless]

I got x = -i squared.

[@christopherellis2663]

X=1. Plainly 😂

[@赖皮球]

1