It so easy to simply inspect and calculate by noticing that 1993^(1/5)~4.5 so you are dealing with very small numbers. Simply notice that 6^5-5^5>1993 so only solutions must be 6
Even simpler my friend: After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993. But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993. If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993. The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction. If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction. Conclusion: there are no solutions. Problem destroyed in 3 minutes.
@@mcwulf25 No, he is not incorrectly making that assumption. Any negative solution will be _balanced_ by a corresponding positive solution. In fact, it is sufficient to examine (6^5 - 5^5) and (5^5 - 4^5).
Here's another way to look at it :) (1) Had an observation: For any integer raised to the power 5, the last digit of the resulting integer is always the same as the last digit of the original integer. (2)Now since the last digit of 1993 is 3, the difference between the last digits of (x,y) will have to be 3. See (1) (3) Therefore (y,x) will have to be in the form of : (0,3) , (1,4) , (2,5) , (3,6) , (4,7) ... (4) The first pair (0,3) gives the value : 3^5 - 0^5 = 243 The second pair (1,4) gives the value : 4^5 - 1^5 = 1023 The third pair (2,5) gives the value 5^5 - 2^5 = 3093 Since 3093>1993 we stop and can safely say that NO SUCH PAIR EXISTS (i.e. NO INTEGER SOLUTION EXISTS) We don't have to go any further as the resulting value of x^5 - y^5 only increases more and more from here.
At 14:43 "they are positive integers". That wasn't a condition of the original problem. Your strategy (as it stands) depends on that extra condition, although it could be modified not to rely on it. Another way of looking at the problem is that x^5 increases very rapidly for successive integers. So fast that 5^5 - 4^5 = 3125 - 1024 is equal to 2101, which is greater than 1993. That means that the absolute value of x and y must be less than 5, since the differences only get larger for larger integers. If any solutions exist, they must be the difference between (or the sum of, if we allow negative integers) the fifth powers of integers from the set {0, 1, 2, 3, 4}, i.e. the difference between numbers from the set {0, 1, 32, 243, 1024}. It doesn't take many moments to realise that you can't make 1993 by combining any two of those.
My solution, which is very simple and destroys the problem in 3 minutes: After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993. But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993. If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993. The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction. If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction. Conclusion: there are no solutions. Problem destroyed in 3 minutes.
Numerically solving is much easier: 1993^(1/5) is between 4 and 5. So the first possible integer solution is 5^5 - 4^5. This value is greater than 1993. There are thus no solutions possible because: n^5 - (n-1)^5 >= 5^5 - 4^5 > 1993 when n >= 5 Hence there is no solution for n > 5. Now consider: n^5 - (n-k)^5 > n^5 - (n-1)^5 >= 5^5 - 4^5 > 1993 when n >= 5 and 1 < k < n This means there are no solutions, at least for positive integers. for negative integers: n^5 + m^5 = 1993 where 1 < n, m < 5 choosing n = 1, 2, 3, 4, does not yield integer value for m. Hence there are no solutions.
Another way to show that x-y can't equal 1 that (to me anyway) seems a bit simpler than expanding the binomial is to realize (several easy ways to do this) that: a^5 = a (mod 5) This gives directly that: x^5-y^5 = x-y (mod 5) Which accomplishes the same thing you did by first expanding the fifth power and then taking mod 5
Here is my solution that is even simpler (your idea can also be used in it): After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993. But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993. If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993. The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction. If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction. Conclusion: there are no solutions. Problem destroyed in 3 minutes.
My reasoning was similar. If we put y=x-1 then the x^5 terms cancel and we subtract 1 to get a polynomial with 5 as a factor equal to 1992. So not possible.
A different solution, imho quicker: case 1. x>0, y>0 then x>1993 and x^5-y^5 > x^5-(x-1)^5. additionally x^5-(x-1)^5 is growing. Therefore only number is 5. Which doesn't do it. Case 2. y0. and is symmetric for x and y, x^5 < 1993. So the x=4,y=4 overshoots. x=4, y=3 must overshoot or be on target. but 4^5=1024 3^5 = 243 so 4^5+3^5 = 1267. No solution then. Last case is x
Hi everyone, i learned on another site something that may be a theorem, that is that (x-y) ALWAYS divides x^n - y^n, so that applying this to our problem tells us that (x-y) divides 1993. but, as 1193 is prime, it cannot be divided and consequently there is no solution.
8:50 The second case is unnecessary since if x-y=1993 then must be x^4+x^3 y+x^2 y^2+x y^3+y^4=1 which has integers solutions ((-1,0),(-1,1),(0,-1),(0,1),(1,-1),(1,0)) and so the difference can't be 1993
The solution becomes simple by exploring the ending digits.... As, 1993 is prime so x = y + 1 [as (x - y) is a factor of (x^5 - y^5) and has to be 1] So x and y are consecutive integers.... With integer and their integer powers, all digits exhibit a cyclicity of 1, 2 or 4.... And an interesting outcome is that the 5th power ALWAYS end with same last digit as the original integer.... Eg. 2^5 = 32, 7^5 = 16807 Which means x^5 and y^5 always end with last digits of x and y and hence, (x^5 - y^5) ends in 1, so can never be equal to 1993 In problems related to Number Theory, especially with Integers, I have always experienced that exploring last digits is a very powerful tool and can simplify solutions to a great extent.
even consecutive quintics are long distances apart, so there are very few candisate numbers to begin with. Moreover, once you establish x-y=k>1 you have x^n-y^n>k, for n>1, because distances between powers increase with the power (for bases outside [0,1]), so the second half of the proof is overkill.
Hey great video thank you ! However Did you need to check the second case ? I feel like if one of the factors must be 1 it is x-y since x>y the second factor won’t give you 1 when x-y is 1993. Also was is restricted to natural numbers ?
My solution, that completely destroys the problem in 3 minutes, uses that idea: After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993. But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993. If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993. The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction. If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction. Conclusion: there are no solutions. Problem destroyed in 3 minutes.
1993 being a prime x= y+1 is the only feasible solution Hereby (y+1)^5 - y^5 = 1993 or 5(y^4 +y) + 10( y^3 + y^2).= 1992 or 5(y+1)y( y^2+y +1) = 1992 Since 5 is not a divisor of 1992, no solution
An interesting extension of the same starting step.... As, 1993 is prime so x = y + 1 [as (x - y) is a factor of (x^5 - y^5) and has to be 1] So x and y are consecutive integers.... With integer and their integer powers, all the digits exhibit a cyclicity of 1, 2 or 4.... And an interesting outcome is that the 5th power ALWAYS end with same last digit as the original integer.... Eg. 2^5 = 32, 7^5 = 16807 Which means x^5 and y^5 always end with last digits of x and y and hence, (x^5 - y^5) ends in 1, so can never be equal to 1993 In problems related to Number Theory, especially with Integers, I have always experienced that exploring last digits is a very powerful tool and can simplify solutions to a great extent.
This problem has in fact no integer solution for x and y, but has real values. I tried in the similar way and found that the approximate solution of x was 4.9403 and of y was 3.9403.
It's unclear to me how you ruled out one being negative. It's easy to see that if both are negative you just switch places and have a symmetric case, but why can't Y be negative and X positive? I can see a few ways to show this but you jumped there faster than i would have and so i assume you have some simple reasoning i didn't quite follow.
@@mathiest good point. You do have to manually check 4^5 + 3^5. Alternatively, as others have indicated, you can notice that 1993, is a prime, and therefore can't be a factor of (a + b).
How I did it x-y = 1 because 1993 is prime and the other factor is too big to be 1 Rewrite problem as x^5 - (x-1)^5 = 1993 Expand: 5x^4 - 10x^3 +10x^2 - 5x + 1 = 1993 5x^4 - 10x^3 +10x^2 - 5x = 1992 5(x^4 - 2x^3 + 2x^2 - x) = 1992 5w = 1992 no integer solutions
Another way I did it....realise 1993 is prime, set u^2 = x etc, difference of squares etc. The addition bracket = 1993, the subtract is 1...no whole number solution.
Yes xy must bi greater than one. Both together can not be one for then (x -y) wolld be zero: Danger. so with x, y greater than one xy must be greater than one. And if some would be negaive the square of xy can not be negative. VERY NICE YOUR ARGUMENT
Substitute y=mx. Then x^5(1-m^5)=1993. So either x^5=1993 or (1-m^5)=1993. None of both cases can be satisfied with integers. So there is no integer solution to this equation.
By the way Sir, kindly try to frame questions that have some integer solutions to these kind of problems so that it would be more interesting to learn. Thank you so much.
I was disappointed with the conclusion. I came up with my own problem that gave a more satisfactory one. Instead of 1,993, I chose 211, also a prime, which happens to be congruent to 1 (mod 5). I knew what the answer would be, but in addition, I got a second solution in which both x and y are negative. I love your channel, but my least favorite problems are ones that turn out to have no solutions, at least among the sought-after type of number (integers, real, etc.). Still, I learn something from all of them.
Why the Hell are you overcomplicating things at such an incredible level????? After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993. But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993. If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993. The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction. If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction. Conclusion: there are no solutions. Problem destroyed in 3 minutes.
You're right. The complex stuff he did makes sense if he doesn't assume x and y are positive. But he says x and y are positive... so there was no need for all that stuff.
I never understood really these diophantic equations. Wha do yo put (y - x) = 1 and not more gneral = a. AHA 1993 IS A PRIMENUMBER AHA. At the end you have somtehing biger than 1, Why is this not good, to be biger than i? Can you explain this. I do not want to be better than you. My concern is manily didactical. Maybe you agree with me, to make a summary at the end of every video. No calculations but only showing the way of solving. Please do not blame me fr this proposition.
@@tonyhaddad1394 Can it be other way of solving?For example x^(5/2)=a and y^(5/2)=b so the equation will be a²-b²=1993 and a and b aslo must be integers. Than a²-b²=(a-b)(a+b)=1993 a-b=1 a+b=1993 2a=1994 a=997 and b=996 Than x^(5/2)=997 and y^(5/2)=996. From this x⁵=997² and y⁵=996². There are no such integers x and y which satisfies this condition.
Im not sure if you allowed to do this substitution but it seems very clean way and you re final answer is right there is no solution but i will think more in you re approache beacaus its very intersting !!!!!!!!
(X+Y)^2(X^2-XY+Y^2)+X^2Y^2 = 1 => X^2 - XY + Y^2 < 0 => X^2 + Y^2 < XY WHICH CONTRADICTS X^2 + Y^2 > 2XY . THANKS. I USED YOUR SEVERAL OF YOUR VIDEOS WHEN TRAINING MY STUDENTS BUT TO BE CREDIBLE I LOOK FOR ALTERNATIVE SOLUTIONS. NOTE THIS ALSO COVERS THE CASE 1993 = (- 1993)(-1)
aha! Ist Modulo 5 eine Zahl gleich drei und ein andere zu vergleichende modulo 5 gleich eins, so können die beiden einander gleich nicht sein. So stellt Sybemath dem PolynomAusdruck elegant das Bein: Zu sagen, da gäbs eine Lösung, das wär gemein. Ein absolut perfektes Argument!--- Ich selber konnte mich nie mit Modulo anfreunden. Er war mir dies immer ein ein Mittagessen voller Kaugummi. Aber Gauss beim 17-Eck braucht auch modulo um grauenhafte Terme zu addieren. Wieso das geht, das kann wohl nicht einmal Sybermath herausfinden. Gauss schreibt einfach Richtiges auf, aber wieso das geht...das weiss Modulo Weisheit niemand?
Easier method to solve this: Solve for integer solutions x^5 - y^5 = 1993 ( x - y ) ( x^4 + x^3 * y + x^2 * y^2 + x * y^3 + y^4 ) = 1993 since 1993 is prime, the only two cases are 1 and 1993 or vice-versa. Case 1 : Let's start with x - y = 1 == > x = y + 1 this means that 5y^4 + 10y^3 + 10y^2 + 5y + 1 = 1993 5y^4 + 10y^3 + 10y^2 + 5y = 1992 5 * ( something ) = 1992 which is impossible Case 2 : x - y = 1993 this means that 1993 * ( x^4 + ........... + y^4 ) = 1993 which means that x^4 + x^3*y + x^2*y^2 + x*y^3 + y^4 = 1 the only way that this can happen is if one of the variables is 0. But if one variable is 0, then you have an irrational solution. Since we are looking for integer solution; the only conclusion is that it is impossible in this case. As a consequence of the work above, there are no integer solutions to the equation x^5 - y^5 = 1993.
By zigmondys theorem there exists a prime p where p|x^5-y^5 ((x,y)=1 because if not then it would be 1993 which means x-y=1) but p doesn t divide x-y since x^5-y^5=1993 which is a prime this means that 1993 doesnt devide x-y this means that x-y=1 using some modular arithmetic we find that can t hold so the equation can t be solved over the integers
Are you not sad, that therre is no soution? And what says the year 1993 itself to it? 1993 says: A whole year I have been waiting for someone to bring me a solution. But noone came to bring it. I am soo sad. I never come again on earth as a full year, how frustration amongst the worldly people to live there. I only asled for two nombers, and noone brougth me them. soooo saaaad this. So EternalBliss Number 1993 went back frustrated to heaven. Then in heven the others asked 1993, how it was. 1993 then sayd: never go on earth, they do not like us, not even two numbers they wanted to give me. THANK YOU THANK YOU, WE THEREFORE NEVER GO DOWN ON EARTH, THE DO NOT LIKE US, 1993 THEN SAYS: BUT DONA NOBIS PACEM THE CAN SING, BUT JUST ONLY SING AS ANENTERTAINMENT. yes yes only entertainment they want
let x=y=a, -> (y+a)^5 - y^5 = 1993 (obviously a > 0) => 5ay(y^3 + 2ay^2 + 2a^2 y+ a^3) = 1993 - a^5 => 1993 mod a == 0, as 1993 is prime, a =1 or a = 1993 (impossible, as the 5ay(y^3 + 2ay^2 + 2a^2 y+ a^3)>0 and 1993- 1993^5 5y(y^3 + 2y^2 + 2y+ 1) = 1992, as 1992 mod 5 != 0, the whole thing is false. So there is no solution.
It so easy to simply inspect and calculate by noticing that 1993^(1/5)~4.5 so you are dealing with very small numbers. Simply notice that 6^5-5^5>1993 so only solutions must be 6
Good insight! Thanks.
Even simpler my friend:
After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993.
But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993.
If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993.
The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction.
If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction.
Conclusion: there are no solutions.
Problem destroyed in 3 minutes.
you are incorrectly assuming the numbers are positive which is kind of in line with this channel that usually assumes the numbers are real :)
Yes that was my approach. Then I realised that x and didn't have to be positive 🙄
@@mcwulf25 No, he is not incorrectly making that assumption. Any negative solution will be _balanced_ by a corresponding positive solution. In fact, it is sufficient to examine (6^5 - 5^5) and (5^5 - 4^5).
Here's another way to look at it :)
(1) Had an observation: For any integer raised to the power 5, the last digit of the resulting integer is always the same as the last digit of the original integer.
(2)Now since the last digit of 1993 is 3, the difference between the last digits of (x,y) will have to be 3. See (1)
(3) Therefore (y,x) will have to be in the form of : (0,3) , (1,4) , (2,5) , (3,6) , (4,7) ...
(4) The first pair (0,3) gives the value : 3^5 - 0^5 = 243
The second pair (1,4) gives the value : 4^5 - 1^5 = 1023
The third pair (2,5) gives the value 5^5 - 2^5 = 3093
Since 3093>1993 we stop and can safely say that NO SUCH PAIR EXISTS
(i.e. NO INTEGER SOLUTION EXISTS)
We don't have to go any further as the resulting value of x^5 - y^5 only increases more and more from here.
Interesting!
Good job. My respects to you.
But what if one of them is negative?
@@aintaintaword666 The original comment described the case where when x>0 & y>0. Lets check for 3 more cases.
case1: x0 & y
At 14:43 "they are positive integers". That wasn't a condition of the original problem. Your strategy (as it stands) depends on that extra condition, although it could be modified not to rely on it.
Another way of looking at the problem is that x^5 increases very rapidly for successive integers. So fast that 5^5 - 4^5 = 3125 - 1024 is equal to 2101, which is greater than 1993. That means that the absolute value of x and y must be less than 5, since the differences only get larger for larger integers.
If any solutions exist, they must be the difference between (or the sum of, if we allow negative integers) the fifth powers of integers from the set {0, 1, 2, 3, 4}, i.e. the difference between numbers from the set {0, 1, 32, 243, 1024}. It doesn't take many moments to realise that you can't make 1993 by combining any two of those.
Nice!
My solution, which is very simple and destroys the problem in 3 minutes:
After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993.
But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993.
If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993.
The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction.
If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction.
Conclusion: there are no solutions.
Problem destroyed in 3 minutes.
I like very much these kind of problem. Awesome resolution. Congratulations.
Glad you like it!
Numerically solving is much easier:
1993^(1/5) is between 4 and 5.
So the first possible integer solution is 5^5 - 4^5.
This value is greater than 1993.
There are thus no solutions possible because:
n^5 - (n-1)^5 >= 5^5 - 4^5 > 1993 when n >= 5
Hence there is no solution for n > 5.
Now consider:
n^5 - (n-k)^5 > n^5 - (n-1)^5 >= 5^5 - 4^5 > 1993 when n >= 5 and 1 < k < n
This means there are no solutions, at least for positive integers.
for negative integers:
n^5 + m^5 = 1993 where 1 < n, m < 5
choosing n = 1, 2, 3, 4, does not yield integer value for m. Hence there are no solutions.
You always pickup the best problems and solve by nice methods. Thankyou for that😊😊
My pleasure 😊
Another way to show that x-y can't equal 1 that (to me anyway) seems a bit simpler than expanding the binomial is to realize (several easy ways to do this) that: a^5 = a (mod 5)
This gives directly that:
x^5-y^5 = x-y (mod 5)
Which accomplishes the same thing you did by first expanding the fifth power and then taking mod 5
Wow! That's so true! Fermat never occurred to me here
Here is my solution that is even simpler (your idea can also be used in it):
After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993.
But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993.
If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993.
The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction.
If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction.
Conclusion: there are no solutions.
Problem destroyed in 3 minutes.
My reasoning was similar. If we put y=x-1 then the x^5 terms cancel and we subtract 1 to get a polynomial with 5 as a factor equal to 1992. So not possible.
A different solution, imho quicker: case 1. x>0, y>0 then x>1993 and x^5-y^5 > x^5-(x-1)^5. additionally x^5-(x-1)^5 is growing. Therefore only number is 5. Which doesn't do it. Case 2. y0. and is symmetric for x and y, x^5 < 1993. So the x=4,y=4 overshoots. x=4, y=3 must overshoot or be on target. but 4^5=1024 3^5 = 243 so 4^5+3^5 = 1267. No solution then. Last case is x
Hi everyone, i learned on another site something that may be a theorem, that is that (x-y) ALWAYS divides x^n - y^n, so that applying this to our problem tells us that (x-y) divides 1993. but, as 1193 is prime, it cannot be divided and consequently there is no solution.
Is it not enough to stop at 10:30? x and why are positive integers, so the expression clearly exceedes 1.
I'm interesting in the resolution of Pell-Fermat ! Thank You !!!
8:50 The second case is unnecessary since if x-y=1993 then must be x^4+x^3 y+x^2 y^2+x y^3+y^4=1 which has integers solutions ((-1,0),(-1,1),(0,-1),(0,1),(1,-1),(1,0)) and so the difference can't be 1993
The solution becomes simple by exploring the ending digits....
As, 1993 is prime so x = y + 1
[as (x - y) is a factor of (x^5 - y^5) and has to be 1]
So x and y are consecutive integers....
With integer and their integer powers, all digits exhibit a cyclicity of 1, 2 or 4.... And an interesting outcome is that the 5th power ALWAYS end with same last digit as the original integer....
Eg. 2^5 = 32, 7^5 = 16807
Which means x^5 and y^5 always end with last digits of x and y and hence,
(x^5 - y^5) ends in 1, so can never be equal to 1993
In problems related to Number Theory, especially with Integers, I have always experienced that exploring last digits is a very powerful tool and can simplify solutions to a great extent.
That was really interesting, and I'm very much a noob when it comes to Diophantine Equations. Thanks so much for sharing!
You're very welcome! 💖
@@SyberMath x=5 ;y=4.081 thanks, Mr: Syber
hey guys does anyone know if syber made any video of factorization metods ?
even consecutive quintics are long distances apart, so there are very few candisate numbers to begin with. Moreover, once you establish x-y=k>1 you have x^n-y^n>k, for n>1, because distances between powers increase with the power (for bases outside [0,1]), so the second half of the proof is overkill.
Awesome video, your explanation has absolute clarity.👍
Great to hear! 😊
Ihre PolynomGleichungen habe ich noch nirgenwo gesehen und auch an der Uni niemals davon gehört. Haben Sie diesen Trick entwickelt?
Hey great video thank you ! However Did you need to check the second case ? I feel like if one of the factors must be 1 it is x-y since x>y the second factor won’t give you 1 when x-y is 1993. Also was is restricted to natural numbers ?
You're right ! I just wanted to show all the cases!
My solution, that completely destroys the problem in 3 minutes, uses that idea:
After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993.
But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993.
If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993.
The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction.
If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction.
Conclusion: there are no solutions.
Problem destroyed in 3 minutes.
1993 being a prime x= y+1 is the only feasible solution
Hereby
(y+1)^5 - y^5 = 1993
or 5(y^4 +y) + 10( y^3 + y^2).= 1992
or 5(y+1)y( y^2+y +1) = 1992
Since 5 is not a divisor of 1992, no solution
An interesting extension of the same starting step....
As, 1993 is prime so x = y + 1
[as (x - y) is a factor of (x^5 - y^5) and has to be 1]
So x and y are consecutive integers....
With integer and their integer powers, all the digits exhibit a cyclicity of 1, 2 or 4.... And an interesting outcome is that the 5th power ALWAYS end with same last digit as the original integer....
Eg. 2^5 = 32, 7^5 = 16807
Which means x^5 and y^5 always end with last digits of x and y and hence,
(x^5 - y^5) ends in 1, so can never be equal to 1993
In problems related to Number Theory, especially with Integers, I have always experienced that exploring last digits is a very powerful tool and can simplify solutions to a great extent.
Fascinating.
Thank you1
What about x-y=-1 and -1993
That cannot be TRUE because
if x-y
This problem has in fact no integer solution for x and y, but has real values. I tried in the similar way and found that the approximate solution of x was 4.9403 and of y was 3.9403.
Once case1 turned out to give no solution, we didn't really have to go for case 2 since (x-y)
your analysis is very clever. thank you.
I appreciate that! 💖
Nice video .Your method of solving is so attractive . You are great .
Thanks a lot
It's unclear to me how you ruled out one being negative. It's easy to see that if both are negative you just switch places and have a symmetric case, but why can't Y be negative and X positive? I can see a few ways to show this but you jumped there faster than i would have and so i assume you have some simple reasoning i didn't quite follow.
If x>0 and y0. What do you mean by negative?
@@SyberMath at 14:45 you claimed that x and y are positive integers. I didn't follow where this claim came from
@@mathiest good point. You do have to manually check 4^5 + 3^5. Alternatively, as others have indicated, you can notice that 1993, is a prime, and therefore can't be a factor of (a + b).
I love your way of solving different problems.thank you.please make more videos.
Thank you, I will
How I did it
x-y = 1 because 1993 is prime and the other factor is too big to be 1
Rewrite problem as x^5 - (x-1)^5 = 1993
Expand: 5x^4 - 10x^3 +10x^2 - 5x + 1 = 1993
5x^4 - 10x^3 +10x^2 - 5x = 1992
5(x^4 - 2x^3 + 2x^2 - x) = 1992
5w = 1992
no integer solutions
I love all the different ways people have come up with, including the one in the video
Nice!
Why is the square of the product of x and y greater than one?
Because x and y are integers
@@SyberMath You are saying x or y cannot be zero and both x and y cannot be one or negative one.
1
2
3
wow you make everything simple and easy
Glad you think so!
Wait what? Can someone explain why xy > 1? It feels like he didn't explain that
Another way I did it....realise 1993 is prime, set u^2 = x etc, difference of squares etc. The addition bracket = 1993, the subtract is 1...no whole number solution.
Good thinking!
Yes xy must bi greater than one. Both together can not be one for then (x -y) wolld be zero: Danger. so with x, y greater than one xy must be greater than one. And if some would be negaive the square of xy can not be negative. VERY NICE YOUR ARGUMENT
Substitute y=mx. Then x^5(1-m^5)=1993. So either x^5=1993 or (1-m^5)=1993. None of both cases can be satisfied with integers. So there is no integer solution to this equation.
By the way Sir, kindly try to frame questions that have some integer solutions to these kind of problems so that it would be more interesting to learn. Thank you so much.
It's important to be able to determine when there are no solutions. It doesn't matter if it's interesting or not
I also would have liked to see this. I created my own offline. Substitute 211 for 1,993.
I was disappointed with the conclusion. I came up with my own problem that gave a more satisfactory one. Instead of 1,993, I chose 211, also a prime, which happens to be congruent to 1 (mod 5). I knew what the answer would be, but in addition, I got a second solution in which both x and y are negative.
I love your channel, but my least favorite problems are ones that turn out to have no solutions, at least among the sought-after type of number (integers, real, etc.). Still, I learn something from all of them.
Thank you for the fair and constructive feedback!
Why the Hell are you overcomplicating things at such an incredible level?????
After the factorisation by (x-y), you notice that if x and y are integers, then x-y and the other barbaric expression are both integers. Then they both divide 1993.
But 1993 is a prime number, and knowing that x>y (since x^5-y^5=1993>0), the only possible values for x-y are 1 or 1993.
If x-y=1 our equation is: (y+1)^5-y^5=1993 => 5.y^4+10^y^3+10y²+5.y+1=1993.
The left member can be written as 5.k+1 where k is an integer. But 1993=5.398+3 so we have a contradiction.
If x-y=1993 that means that the barbaric expression must be equal to 1. But it's > x^4 which is > 1993^4. Another contradiction.
Conclusion: there are no solutions.
Problem destroyed in 3 minutes.
You're right. The complex stuff he did makes sense if he doesn't assume x and y are positive. But he says x and y are positive... so there was no need for all that stuff.
what is your name & where are you from? 😀
I never understood really these diophantic equations. Wha do yo put (y - x) = 1 and not more gneral = a. AHA 1993 IS A PRIMENUMBER AHA. At the end you have somtehing biger than 1, Why is this not good, to be biger than i? Can you explain this. I do not want to be better than you. My concern is manily didactical. Maybe you agree with me, to make a summary at the end of every video. No calculations but only showing the way of solving. Please do not blame me fr this proposition.
Are x and y natural numbers?
No x and y integers ,means it can be negative
@@tonyhaddad1394 Ok, thanks.
@@tonyhaddad1394 Can it be other way of solving?For example x^(5/2)=a and y^(5/2)=b so the equation will be a²-b²=1993 and a and b aslo must be integers. Than
a²-b²=(a-b)(a+b)=1993
a-b=1
a+b=1993
2a=1994
a=997 and b=996
Than x^(5/2)=997 and y^(5/2)=996. From this x⁵=997² and y⁵=996². There are no such integers x and y which satisfies this condition.
Im not sure if you allowed to do this substitution but it seems very clean way and you re final answer is right there is no solution but i will think more in you re approache beacaus its very intersting !!!!!!!!
@@tonyhaddad1394 Ok, I'm waiting for channel owner to respond. Thx
Let p1, p2.....pk be distinct prime numbers. Prove that the equation
X1^p1+x2^p2+....x(k-1)^p(k-1)=xk^pk
has infinitely many natural solutions
I sent you an equ to think about it it's
X^(a)+Y^(b)=857 ,X,Y,a,b are variables and not equal to 1 or 0
I tried for several ways and the result was always no solution! I just couldn't believe that was the solution to the exercise.
Well done! 😁
(X+Y)^2(X^2-XY+Y^2)+X^2Y^2 = 1 => X^2 - XY + Y^2 < 0 => X^2 + Y^2 < XY
WHICH CONTRADICTS X^2 + Y^2 > 2XY . THANKS. I USED YOUR SEVERAL OF YOUR VIDEOS WHEN TRAINING MY STUDENTS BUT TO BE CREDIBLE I LOOK FOR ALTERNATIVE SOLUTIONS.
NOTE THIS ALSO COVERS THE CASE 1993 = (- 1993)(-1)
You are a teacher? How nice! You're very welcome
aha! Ist Modulo 5 eine Zahl gleich drei und ein andere zu vergleichende modulo 5 gleich eins, so können die beiden einander gleich nicht sein. So stellt Sybemath dem PolynomAusdruck elegant das Bein: Zu sagen, da gäbs eine Lösung, das wär gemein. Ein absolut perfektes Argument!--- Ich selber konnte mich nie mit Modulo anfreunden. Er war mir dies immer ein ein Mittagessen voller Kaugummi. Aber Gauss beim 17-Eck braucht auch modulo um grauenhafte Terme zu addieren. Wieso das geht, das kann wohl nicht einmal Sybermath herausfinden. Gauss schreibt einfach Richtiges auf, aber wieso das geht...das weiss Modulo Weisheit niemand?
Easier method to solve this:
Solve for integer solutions
x^5 - y^5 = 1993
( x - y ) ( x^4 + x^3 * y + x^2 * y^2 + x * y^3 + y^4 ) = 1993
since 1993 is prime, the only two cases are 1 and 1993 or vice-versa.
Case 1 : Let's start with x - y = 1 == > x = y + 1
this means that 5y^4 + 10y^3 + 10y^2 + 5y + 1 = 1993
5y^4 + 10y^3 + 10y^2 + 5y = 1992
5 * ( something ) = 1992 which is impossible
Case 2 : x - y = 1993
this means that 1993 * ( x^4 + ........... + y^4 ) = 1993
which means that x^4 + x^3*y + x^2*y^2 + x*y^3 + y^4 = 1
the only way that this can happen is if one of the variables is 0. But if one variable is 0, then you have an irrational solution. Since we are looking for integer solution; the only conclusion is that it is impossible in this case.
As a consequence of the work above, there are no integer solutions to the equation x^5 - y^5 = 1993.
Nice! I like your approach.
x=(2×1993)^(1/5) , y= 1993^(1/5) can be a solution of the equation
INTEGRAL SOLN is asked
Superb
Thanks 🤗
If I answer this in the exam as "NO ANSWER"
do I get a credit, since it really doesn't have any solution..
You should
@@SyberMath you are the best!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Very interesting.
Glad you think so!
In about 5 minutes I got X=5.5185 and Y=5
By zigmondys theorem there exists a prime p where p|x^5-y^5 ((x,y)=1 because if not then it would be 1993 which means x-y=1) but p doesn t divide x-y since x^5-y^5=1993 which is a prime this means that 1993 doesnt devide x-y this means that x-y=1 using some modular arithmetic we find that can t hold so the equation can t be solved over the integers
1993|37
185 54
143
148
9
Нет результатов. Вот что значит надеяться на похожее в таких уравнениях.
No solution i get it weeee!!!!!😀
Yay!!! 😊
@@SyberMath 💓💓💓
❤❤❤
5X-5y=1993
0XY=1993/0=0
Xy=0
Are you not sad, that therre is no soution? And what says the year 1993 itself to it? 1993 says: A whole year I have been waiting for someone to bring me a solution. But noone came to bring it. I am soo sad. I never come again on earth as a full year, how frustration amongst the worldly people to live there. I only asled for two nombers, and noone brougth me them. soooo saaaad this. So EternalBliss Number 1993 went back frustrated to heaven. Then in heven the others asked 1993, how it was. 1993 then sayd: never go on earth, they do not like us, not even two numbers they wanted to give me. THANK YOU THANK YOU, WE THEREFORE NEVER GO DOWN ON EARTH, THE DO NOT LIKE US, 1993 THEN SAYS: BUT DONA NOBIS PACEM THE CAN SING, BUT JUST ONLY SING AS ANENTERTAINMENT. yes yes only entertainment they want
தமிழ் மொழி ஐயா 👍🏻✍️
J aime le bon sens 👌
let x=y=a, -> (y+a)^5 - y^5 = 1993 (obviously a > 0)
=> 5ay(y^3 + 2ay^2 + 2a^2 y+ a^3) = 1993 - a^5
=> 1993 mod a == 0, as 1993 is prime, a =1 or a = 1993 (impossible, as the 5ay(y^3 + 2ay^2 + 2a^2 y+ a^3)>0 and 1993- 1993^5 5y(y^3 + 2y^2 + 2y+ 1) = 1992, as 1992 mod 5 != 0, the whole thing is false.
So there is no solution.
X-y|1993
What is the solution?
Replace 1993 by 2101 in the year 2101 and ask the same question.😄😄 Then we have a solution
Make explanations a bit shorter & understandable. Mathematics - otherwise -would seem a boring exercise
الترجمة للعربية اذا امكن من فضلك
إنه باللغة الإنجليزية
@@SyberMath ترجمة الفيديو للعربية اذا امكن؟
Nice problem !!!!!!!
Thank you!
and fun
x^y-y^x=17.
X-?.Y-?
👍
Disappointing video. Normally they are good but to watch all this for no solutions put me off
There is a solution, (1994^1/5)^5 - 1^5 = 1993, then x= 4,57, y = 1
Integers?