Nice problem. Note that, towards the beginning of the proof, you established that x>2y. Thus, k>2. Therefore, there was no need to test the case that k=2 at the very end of the proof.
But x isnt > 2y, because if you choose x =3 and y=2 then x is not greater than 2y=4, and another example is if x =4 and y=3, again x=4 isnt greater than 3 times 2=6
The values that you gave are not solutions to the system. In the case that x and y are not equal, it was clearly established that x>2y. I recommend that you rewatch the video and write down the solutions for x and y (not k).
4:18 I don't understand this step, could someone clarify: if x/y > 1 and x/2y >1 therefore x >2y, then how is x/y an integer (more specifically y|x)? For example this does not hold if I take the values x=5 and y=2
I think the logic is like this. At that step, we know that the RHS y^(x-2y) is guaranteed to be an integer. The LHS (x/y)^y is also an integer. If x is not a multiple of y, the LHS is a fraction, and will not be an integer. Therefore x must be a multiple of y.
x/y > 1 means y^(x-2y) must also be > 1, this means x-2y > 0, x > 2y. And because the RHS (y^(x-2y)) is an integer the LHS must be an integer as well, and (x/y)^y can only be an integer if x/y is an integer
It’s early morning when I’m watching this, so I’m sure I’m missing something, but I don’t understand the reduction that was done with k to the y equals y to the y time k-2 going to k equals y to the k-2. That implies y to the y equals k to the y-1 and I just don’t see where that came from.
Here's my somewhat wonky solution: (xy)^y = y^x x^y = y^(x-y) From this we can conclude two things: x and y must contain the same factors, and x>y. Not 100% sure on how to prove that these MUST be true, but it feels very intuitive.. This means x = y^n y^(yn) = y^(y^n - y) yn = y^n - y y(n+1) = y^n n+1 = y^(n-1) lg(n+1) = (n-1)*lgy y = 10^(lg(n+1)/(n-1)) = (n+1)^(1/(n-1)) or the (n-1)th root of (n+1) From this find all integer n such that y is an integer. I suppose you can do this by hand and checking various limits, but I just checked the graph and concluded n = 1, n = 2 and n = 3 are the only solutions. This gives the three solutions (1, 1), (9, 3) and (8, 2).
Your second step of substituting x=y^n is incorrect. While it is true that the prime factors of both numbers have to be identical, it is not proven that x must be an integer power of y. Consider the hypothetical case x=8m, y=4n. From the conclusions we’ve reached yet it is not impossible that m and n are selected such that x is not an integer power of y, and yet the RHS is equal to it
@@armacham exactly, in my opinion when you raise a number by the 0 power, it is the same to say x^0 = x/x, so for example 3^0 = 3/3 = 1, but for 0^0 = 0/0 = undefined. That’s just how I like see it ;)
If x > y how is x > 2y for all integers, for example if x is 3 and y is 2 then x < 2y because 3 < 2 times 2 so its a contradiction, x is not larger than 2y
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Nice problem. Note that, towards the beginning of the proof, you established that x>2y. Thus, k>2. Therefore, there was no need to test the case that k=2 at the very end of the proof.
But x isnt > 2y, because if you choose x =3 and y=2 then x is not greater than 2y=4, and another example is if x =4 and y=3, again x=4 isnt greater than 3 times 2=6
The values that you gave are not solutions to the system. In the case that x and y are not equal, it was clearly established that x>2y. I recommend that you rewatch the video and write down the solutions for x and y (not k).
@@wkbj7924 right, so as well as assuming that x>y we are also assuming that x>2y, so we have a case within a case?
@@thehardlife5588 No. Towards the very beginning of the case in which x>y, it is proven via rules of exponents that x>2y.
I established x = y^k instead of x=k*y , led to similar results anyway.
In k = y^(k-2) clearly a smaller value of y would permit a larger solution for k, so it makes sense you treated y=2 in your proof k
Can you solve the following equation in natural numbers: x^2 + y^2 = 5xy +7 ?? Thanks
4:18 I don't understand this step, could someone clarify:
if x/y > 1 and x/2y >1 therefore x >2y, then how is x/y an integer (more specifically y|x)? For example this does not hold if I take the values x=5 and y=2
I think the logic is like this.
At that step, we know that the RHS y^(x-2y) is guaranteed to be an integer. The LHS (x/y)^y is also an integer. If x is not a multiple of y, the LHS is a fraction, and will not be an integer.
Therefore x must be a multiple of y.
x/y > 1 means y^(x-2y) must also be > 1, this means x-2y > 0, x > 2y. And because the RHS (y^(x-2y)) is an integer the LHS must be an integer as well, and (x/y)^y can only be an integer if x/y is an integer
Nice knowledge...
It’s early morning when I’m watching this, so I’m sure I’m missing something, but I don’t understand the reduction that was done with k to the y equals y to the y time k-2 going to k equals y to the k-2. That implies y to the y equals k to the y-1 and I just don’t see where that came from.
You raise both sides to 1/y power or equivalently take the y'th root on both sides. And you can do this, because y is a positive integer.
Here's my somewhat wonky solution:
(xy)^y = y^x
x^y = y^(x-y)
From this we can conclude two things: x and y must contain the same factors, and x>y. Not 100% sure on how to prove that these MUST be true, but it feels very intuitive..
This means x = y^n
y^(yn) = y^(y^n - y)
yn = y^n - y
y(n+1) = y^n
n+1 = y^(n-1)
lg(n+1) = (n-1)*lgy
y = 10^(lg(n+1)/(n-1)) = (n+1)^(1/(n-1)) or the (n-1)th root of (n+1)
From this find all integer n such that y is an integer. I suppose you can do this by hand and checking various limits, but I just checked the graph and concluded n = 1, n = 2 and n = 3 are the only solutions.
This gives the three solutions (1, 1), (9, 3) and (8, 2).
Your second step of substituting x=y^n is incorrect. While it is true that the prime factors of both numbers have to be identical, it is not proven that x must be an integer power of y. Consider the hypothetical case x=8m, y=4n. From the conclusions we’ve reached yet it is not impossible that m and n are selected such that x is not an integer power of y, and yet the RHS is equal to it
What about 1=y, x=2?
Shouldn't x>=2y? Not x>2y?
Sorry, but at 7:04, why do you maintain that for y>=2, k>=2^(k-2)?
isn't 1 trivially testable?
Isn't x=0 and y=0 also a possible solution for the first case?
He assumed that x and y are positive integers 0:00
@@sniegsnygg ah yeah of course thx
@@sniegsnygg Additionally, 0^0 is undefined. If you plug in 0s, you get: "undefined = undefined". So that is not considered to be true.
@@armacham exactly, in my opinion when you raise a number by the 0 power, it is the same to say x^0 = x/x, so for example 3^0 = 3/3 = 1, but for 0^0 = 0/0 = undefined. That’s just how I like see it ;)
@@armacham 0^0 as a limit is undefined, but 0^0 as a number is 1.
nice!
you can write it as x/y = lnx/lny + 1
Sure, but what would you do then?
Nice problem. Kind of how I solved it but not as rigourous as you!
If x > y how is x > 2y for all integers, for example if x is 3 and y is 2 then x < 2y because 3 < 2 times 2 so its a contradiction, x is not larger than 2y
Did you miss the whole part where he explained that if x>y then the LHS>1 and thus the exponent on the RHS>0? If x-2y>0, then x>2y, IN THIS PROBLEM.
@@bubbletea-ol4lr yes i got it
asnwer= y isit
(1×1)¹=1¹
This was equ fanction because impress integer numbers.
prove power.
log(xy) y = log(y) x
log(y) y * log(x) y = log(y) x
1/log(y) y max often
log(x) y =x/y Answer
inlaw mass F = ma
F=log(x) y
ma x/y