A Quintic Diophantine Equation (x^5-y^5=1993)

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[sawyerw]

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<, and then just calculate 1, 2, 3, 4, 5 to 5th power and see that there are no solutions adding or subtracting. It would be better problem to use much bigger numbers forcing the use of your technique (which is nicely instructive).

[palindrame]

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.

[RexxSchneider]

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.

[rafael]

I like very much these kind of problem. Awesome resolution. Congratulations.

[sahilsinghbhandari]

You always pickup the best problems and solve by nice methods. Thankyou for that😊😊

[PunmasterSTP]

That was really interesting, and I'm very much a noob when it comes to Diophantine Equations. Thanks so much for sharing!

[bollyfan]

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.

[yhamainjohn]

I'm interesting in the resolution of Pell-Fermat ! Thank You !!!

[damiangruszka]

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. y<0. Then x^5-y^5 can be written as x^5+y^5, y>0. 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<0, then y <0 as well. So we can write it as y^5-x^5=1993, x, y>0, which we handled. Ergo no solutions. Writing this was longer than solving in the head.

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.

[mathiest]

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

[sonaraghavan]

Awesome video, your explanation has absolute clarity.👍

[balthazarbeutelwolf]

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.

[christianthomas]

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.

[-sn]

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 ?

[lazymello]

Once case1 turned out to give no solution, we didn't really have to go for case 2 since (x-y)<(the other factor), so x-y can't be 1993 while the other factor is 1

[Caturiya]

Ihre PolynomGleichungen habe ich noch nirgenwo gesehen und auch an der Uni niemals davon gehört. Haben Sie diesen Trick entwickelt?

[mfgjvalkon]

hey guys does anyone know if syber made any video of factorization metods ?

[ramaprasadghosh]

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

[aliasgharheidaritabar]

I love your way of solving different problems.thank you.please make more videos.