I went for all sol-s, it's *BRUTAL* . I raised 1st to 5th and 2nd to 3rd power to homogenize it and get 5t⁹-3t⁷+10t⁶+10t³-3t²+5=0 for t=x/y, with t=-1 a root, divide: 5t⁸-5t⁷+2t⁶+8t⁵-8t⁴+8t³+2t²-5t+5, set z=t+1/t: 5z⁴-5z³-18z²+23z-2=0 with z=-2 a root, divide: 5z³-15z²+12z-1=0, set z-1=u: 5u³-3u+1=0 which we can finally solve (Cardano). P.S. This (obviously) means much easier time for real roots only because, first, t=-1 and z=-2 mean the same because z=t+t/t=-2 only leads to t=-1. But t=-1 yields x=-y which is impossible and - second - we don't have to deal with (brutal) cubic roots because we can show that for the f(u) = 5u³-3u+1 we get f'=15u²-3, so u=±√5/5 are critical points, checking the sign gets f(max)=f(-√5/5), f(min)=f(√5/5) with f increasing on (-∞, -√5/5) and (√5/5, +∞) while decreasing on (-√5/5, √5/5). And since f(√5/5) = (5-2√5)/5 > 0 there can be no real roots on (-√5/5, +∞) meaning the real root is on (-∞, -√5/5), but there f monotonously increases so the root is single. Easy check shows f(-1) = -5+3+1 = -1 < 0, so our root u₀ ∈ (-1, -√5/5), but that's in u meaning z₀ ∈ (0, 1-√5/5). However |t+t/t| ≥ 2 for t∈ℝ so finally t+t/t = z₀ cannot have any real roots. EDIT: Forgot to mention that on the first step (making it homogeneous) we will arrive at "degree 15" equation but with the factor of x³y³ which we can then set to 0. That's from where we get the nonic polynomial and also from that x³y³ we get the only possible real solutions (1, 0) and (0, 1)
Bezout's theorem ( consider points of infinity are solution the equation {x^3+y^3=0, x^5+y^5=0}. Bezout's theorem applies to x^3+y^3=z^3, x^5+y^5=z^5. (Points at infinity when z=0). Point at infinity are x+y=0,z=0 (1:-1:0) (we can prove that multiplicity is 3), Other points at infinity are solution of {x^2-x*y+y^2=0, x^4-x^3*y+x^2*y^2-x*y^3+y^4=0, x ot=0 and y ot=0.} =\ emptyset. By Bezout's theorem there are 15 points. Edit: we can prove that (0:1:1) and (1:0:1) (the real points) have multiplicity 3. Other 6 points (no real) have multiplicity 1.
When both x and y are non negative, the two equations basically say that the L_3 distance from (x, y) from the origin is the same as L_5 distance. But L_p norm decreases as p increases, unless x or y is 0. If, say, x < 0, then y must be > 0. Let u=1/y and v=-x/y, and we get u^3 + v^3 = 1 and u^5 + v^5 = 1 with both u and v >0. So we are back in the previous case.
Easier solution: Consider shape of each as function: y^3 = (1-x^3) y^5 = (1-x^5) Since y=y, just consider when can y^3 = y^5. Obvious: only if y=0,y=1,y=-1 This is obvious because of shape of parent odd function. Absolute value between 0,1 smaller for bigger power. Absolute value above 1 bigger for bigger power. Test each value. From list of candidate: y=0? 0 + x^3 = 1 0 + x^5 = 1 1^(1/3) = 1^(1/5) -> x= 1 (1,0)✅ y=1? 1 + x^3 = 1 1 + x^5 = 1 0^(1/3) = 0^(1/5) -> x=0 (0,1)✅ y=-1? -1 + x^3 = 1 -1 + x^5 = 1 x^3 = 2 x^5 = 2 2^(1/3) ≠ 2^(1/5)❌ Answer all cases of ✅: (0,1) (1,0)
@@TheWarmestWaffle I know that they aren't equal but which case / idea makes you think that y^3 = y^5? Both equations have different value and I still don't get it how y^3 = y^5
@@rifkiarza6590 you just need to find all the possible intersections on one axis and then plug in and test the other. You could compare either the LHS of both or the RHS of both. Putting it in that form makes it easy to tell is all. :)
I reckoned this from a geometrical approach too. Then I think the answers (0,1) and (1,0) apply to all pairs of equations of this form with any odd positive power.
let X=x³,then y³=1-X, then x⁵+y⁵=X⁵÷³+(1-X)⁵÷³， by the monotonous of the fuction X⁵÷³+(1-X)⁵÷³ when X≥-1/2 and the function is symmetrical at X=-1/2. We can see the pair(X,Y)=(0,1) and (1,0) are the only solutions. By this way we can change the 5 in the condition to7, 9, 11…… and get all the same solution.
This is very easy to see the real sol are only(1,0)or(0,1) the real problem is to find the complex solutions! I enjoyed watching all of your solution,but i did not like this problem
@@hareal3904 subtract the 2 eq and you get:( x^3-x^5)+(y^3-y^5)=0 now define f(x)=x^3-x^5 so we have f(x)+f(y)=0 so there are 2 cases: f(x) and f(y) are both 0 this is when x=-1,0,1 x=-1 no solutions x=0 y=1 solution and x=1 y=0 is asolution. The other case is f(x)=-f(y). Synce f is odd y=-x,but y=-x do not give you asolution. (If you draw f(x)=x^3-x^5 you will see that the function is not one to one only for-1
@@hareal3904 x^3+y^3=1 and x^5+y^5=1 are a pair of shapes similar to a squircle, but with different closeness to squares. The fifth-power one bulges more than the third-power one everywhere but the the x- and y-intercepts.
Always odd power equtions has at least one solution, we can see that when we calculat limits at ±∞, from -∞ to +∞ ther is defenetly an intersection point
It happened. Math videos have swarmed my recommended page
Wow, this proof by showing P>0 to get to the solution is just elegant as always!!!
A graph seems a good way to find solutions
x^5 - x^3 divided into six regions, boundaries at f(x) = 0 or f(x) = f(y) for some y where f'(y) = 0 but x =/= y. then proof by casework
I went for all sol-s, it's *BRUTAL* . I raised 1st to 5th and 2nd to 3rd power to homogenize it and get 5t⁹-3t⁷+10t⁶+10t³-3t²+5=0 for t=x/y, with t=-1 a root, divide: 5t⁸-5t⁷+2t⁶+8t⁵-8t⁴+8t³+2t²-5t+5, set z=t+1/t: 5z⁴-5z³-18z²+23z-2=0 with z=-2 a root, divide: 5z³-15z²+12z-1=0, set z-1=u: 5u³-3u+1=0 which we can finally solve (Cardano).
P.S. This (obviously) means much easier time for real roots only because, first, t=-1 and z=-2 mean the same because z=t+t/t=-2 only leads to t=-1. But t=-1 yields x=-y which is impossible and - second - we don't have to deal with (brutal) cubic roots because we can show that for the f(u) = 5u³-3u+1 we get f'=15u²-3, so u=±√5/5 are critical points, checking the sign gets f(max)=f(-√5/5), f(min)=f(√5/5) with f increasing on (-∞, -√5/5) and (√5/5, +∞) while decreasing on (-√5/5, √5/5). And since f(√5/5) = (5-2√5)/5 > 0 there can be no real roots on (-√5/5, +∞) meaning the real root is on (-∞, -√5/5), but there f monotonously increases so the root is single. Easy check shows f(-1) = -5+3+1 = -1 < 0, so our root u₀ ∈ (-1, -√5/5), but that's in u meaning z₀ ∈ (0, 1-√5/5). However |t+t/t| ≥ 2 for t∈ℝ so finally t+t/t = z₀ cannot have any real roots.
EDIT: Forgot to mention that on the first step (making it homogeneous) we will arrive at "degree 15" equation but with the factor of x³y³ which we can then set to 0. That's from where we get the nonic polynomial and also from that x³y³ we get the only possible real solutions (1, 0) and (0, 1)
Well done!
I love your solution. Would love to subscribe if you tell me you also have a math youtube channel hahaha
Bezout's theorem ( consider points of infinity are solution the equation {x^3+y^3=0, x^5+y^5=0}. Bezout's theorem applies to x^3+y^3=z^3, x^5+y^5=z^5. (Points at infinity when z=0). Point at infinity are x+y=0,z=0 (1:-1:0) (we can prove that multiplicity is 3), Other points at infinity are solution of {x^2-x*y+y^2=0, x^4-x^3*y+x^2*y^2-x*y^3+y^4=0, x
ot=0 and y
ot=0.} =\ emptyset. By Bezout's theorem there are 15 points.
Edit: we can prove that (0:1:1) and (1:0:1) (the real points) have multiplicity 3. Other 6 points (no real) have multiplicity 1.
The solution x=-y is a solution of infinity (z=0). Therefore this solution is solution the resultant.
@@tianqilong8366 Math channel? Ha, as if! I'm just an amateur and never had a formal math training, it's all just for fun :)
Fantastic! Thanks for the great clear explanations!
When both x and y are non negative, the two equations basically say that the L_3 distance from (x, y) from the origin is the same as L_5 distance. But L_p norm decreases as p increases, unless x or y is 0.
If, say, x < 0, then y must be > 0. Let u=1/y and v=-x/y, and we get u^3 + v^3 = 1 and u^5 + v^5 = 1 with both u and v >0. So we are back in the previous case.
Easier solution:
Consider shape of each as function:
y^3 = (1-x^3)
y^5 = (1-x^5)
Since y=y, just consider when can y^3 = y^5.
Obvious: only if y=0,y=1,y=-1
This is obvious because of shape of parent odd function. Absolute value between 0,1 smaller for bigger power. Absolute value above 1 bigger for bigger power.
Test each value. From list of candidate:
y=0?
0 + x^3 = 1
0 + x^5 = 1
1^(1/3) = 1^(1/5)
-> x= 1
(1,0)✅
y=1?
1 + x^3 = 1
1 + x^5 = 1
0^(1/3) = 0^(1/5)
-> x=0
(0,1)✅
y=-1?
-1 + x^3 = 1
-1 + x^5 = 1
x^3 = 2
x^5 = 2
2^(1/3) ≠ 2^(1/5)❌
Answer all cases of ✅: (0,1) (1,0)
that's cool but I still don't get it why y^3 = y^5
@@rifkiarza6590 it’s not that they’re equal, it’s that you only need to care about when they could be equal.
@@TheWarmestWaffle I know that they aren't equal but which case / idea makes you think that y^3 = y^5? Both equations have different value and I still don't get it how y^3 = y^5
@@rifkiarza6590 you just need to find all the possible intersections on one axis and then plug in and test the other. You could compare either the LHS of both or the RHS of both. Putting it in that form makes it easy to tell is all. :)
I reckoned this from a geometrical approach too. Then I think the answers (0,1) and (1,0) apply to all pairs of equations of this form with any odd positive power.
let X=x³,then y³=1-X, then x⁵+y⁵=X⁵÷³+(1-X)⁵÷³， by the monotonous of the fuction X⁵÷³+(1-X)⁵÷³ when X≥-1/2 and the function is symmetrical at X=-1/2. We can see the pair(X,Y)=(0,1) and (1,0) are the only solutions. By this way we can change the 5 in the condition to7, 9, 11…… and get all the same solution.
When he transformed the equation adding fraction I understand I was lost
p^3+3p^2+6p+5 right? Not 6p^3
Yes, I wrote an extra cube there. Thank you for spotting that.
I did this in one second
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This is very easy to see the real sol are only(1,0)or(0,1) the real problem is to find the complex solutions! I enjoyed watching all of your solution,but i did not like this problem
How is it so easy to see that (1,0) (0,1) are the only real solutions?
@@hareal3904 subtract the 2 eq and you get:( x^3-x^5)+(y^3-y^5)=0 now define f(x)=x^3-x^5 so we have f(x)+f(y)=0 so there are 2 cases: f(x) and f(y) are both 0 this is when x=-1,0,1 x=-1 no solutions x=0 y=1 solution and x=1 y=0 is asolution. The other case is f(x)=-f(y). Synce f is odd y=-x,but y=-x do not give you asolution. (If you draw f(x)=x^3-x^5 you will see that the function is not one to one only for-1
@@hareal3904 x^3+y^3=1 and x^5+y^5=1 are a pair of shapes similar to a squircle, but with different closeness to squares. The fifth-power one bulges more than the third-power one everywhere but the the x- and y-intercepts.
@@yoav613 f being odd doesnt make this implication true:
f(x) = -f(y) => x=-y
We can only say the other way around is true:
x=-y => f(x) = -f(y)
@@LastHopeee when f is also one to one it dose. That is why i mentiond that in the region -1
Namaste🙏.
Always odd power equtions has at least one solution, we can see that when we calculat limits at ±∞, from -∞ to +∞ ther is defenetly an intersection point
You must say x^2-xy+y^2 is not zero to divide by it.
You forgot to mention that p can't be zero. Nice video anyways.