Solving a higher degree Diophantine equation

[@bot24032]

11:45 no, that would just be z=-1 in our branch. in fact, the change z->-z-3 does not change the formula for y, meaning it is a bijection between branches (every solution can be presented as a solution of whichever branch you wish)

[@gianluca4893]

If there Is a bijection then the solutions Will be the same?

[@bot24032]

@@gianluca4893 in the case when it is a change of variables that doesn't change the equations (that includes, for example, the fact that z is still an integer), then yes. In our case, you could say (if you want to be more rigorous) "for the solutions of the second branch, let z=-3-z_1, then z_1 is an integer, y=(-3-z_1)(-3-z_1+3)=z_1(z_1+3) and x³=-(-3-z_1)(-3-z_1+2)(-3-z_1+3)=z_1(z_1+1)(z_1+3). That is the form of a solution of the first branch with z=z_1, meaning we already covered that case." Hope this helped!

[@Alex_Deam]

@@bot24032 Yeah, at this point we're assuming x=z, so if you start off by plugging in x=0 to get y=-2, the z value derived from that is no longer equal to x, so no solutions were missed as it's a bijection as you say

[@gianluca4893]

@@bot24032 Wow, thank you so much! Great explanation! Do you compete in math olympiad?

[@bot24032]

@@gianluca4893 I was competing in some smaller olympiads and was planning to do the national one, but have to move countries and to compete now I would have to a) know German well enough to write solutions in it and b) have a good understanding of what is and isn't allowed to be used without proof in my solutions (I don't even know the school curriculum, so I'm never sure if I even can use what I think of as basic)

[@confusedsoul4775]

considering the given eqn as quadratic in x^3, and applying the condition that the discriminant is a perfect square,we get 4y+9 = k^2 for some k€ N. putting this value of y in the given eqn and solving for x^3, we get -x^3= (k^2-9)(1+ k)/8 or -x^3 =(k^2-9)(1-k)/8 thus, we must have (k^2-9)(1+k) is a perfect cube or (k^2-9)(1-k) is a perfect cube in case 1, if (k^2-9)(1+k) =c^3 for some integer c. clearly,k =3 gives c=0 as a solution for values of k>9, we get k^2- 9k -9 =(c-k)(c^2 +ck + k^2) for values of k>9, we have l.hs of above eqn and hence, r.hs are both positive integers. thus, 0

[@rhysjones1432]

at 15:26 the quadratic factor was y^2 +6y+16, not y^2 +6y^2+16

[@gennarobullo89]

I love these kind of problems

[@tracyh5751]

13:07 by the trichotomy property of integers, we are done. There is no need to consider x = z+1.

[@RexxSchneider]

I'm afraid we have to consider x = z+1, particularly as it leads to the solution x=2, y=4. We are not looking at linear relationships, but actually at quadratic ones, so as x increases from large and negative to large and positive, the value of the expression (x^3)^2 + y(x^3) - (y^3 + 2y^2) decreases from large and positive, through a minimum, and increases back to large and positive again. The value of y determines if the expression is ever negative, and therefore would have an opportunity to attain a value of 0, but that might occur twice, before or after the minimum. You'll find that y has to be greater than -3 for the possibility of the expression reaching zero to exist. Trichotomy isn't at play here.

[@MyOneFiftiethOfADollar]

Huh, you must have been in the mood to pronounce 4 syllable words. Explain more carefully why a>b, a=b, a

[@DOROnoDORO]

day 3 of asking Michael what brand chalk he uses

[@soyoltoi]

Hagoromo

[@DOROnoDORO]

@@soyoltoi does hagoromo sell a 10 color version?

[@tomkerruish2982]

​@@soyoltoiJust like Papa Flammy?

[@hagenfarrell]

More than likely Hagoromo, it’s what most mathematicians use; as do I. It’s wonderful chalk!

[@DOROnoDORO]

@@hagenfarrell Are these 10 colors standard for colored chalk? I only ask because I bought some pretty good colored chalk (not hagoromo, unfortunately) and it's the *exact* same colors

[@siddharthchabra9022]

I mean how do you proove that there is no non integer z which yields and integer x and integer y

[@fartoxedm5638]

By the way this equation defines an elliptic curve over the real numbers!

[@gp-ht7ug]

20:07

[@charleyhoward4594]

this guy is indefatigable ...

[@Francisco-vl5ub]

18:11 (z+3)^3

[@egillandersson1780]

Why my subscription to your channel suddenly disappeared ? I subscribed again, so ok now.

[@emanuellandeholm5657]

Minor quibble, the empty set is an element of the empty set. The problem is the empty set is not in Z.

[@Czeckie]

I hate these pedantic comments that add nothing. And it's even worse when they are wrong. Empty set doesn't have any elements. It's irrelevant tho.

[@ilyashick3178]

interested to see belarus source of solution as well.

[@pietropellegrini7926]

18:13 I think the cube is wrong (should have been z^3 + 3z^2 + 9z +27)

[@donach9]

No, you pick up a binomial 3 and a 3 from the constant term to give you the 9

[@ВасилийЗайцев-н9ю]

Missing derivation of the fact, that z is integer

[@Fred-yq3fs]

Review the video again from 2:45. 4y+9 must be equal to t^2 where t is in an integer (so that the square root yields an integer). Such a t must be odd 'cause 4y+9 is odd, so we might as well write t=2z+3 where z is an integer.

[@MichalZavadil-i4j]

Mistake in proof? The 'z' variable is not intiger by definition. Becouse 2*z+3 = k^2 not implicated it. E.g. z=1/2 is valid not integer value.

[@eiseks3410]

You areing correctful

[@meruempro]

If z is not an integer but 2z+3 is then 2z+3 is even which is not true by definition