Nice problem! You can avoid the quadratic by making two substitutions u = a - b v = a+ b We have (u^2 + v^2)/2 = u^3 Rearranging v^2 = u^2(2u -1) Therefore u divides v. Put v = ku k^2 u^2 = u^2(2u - 1) k^2 = 2u -1 u = (k^2 + 1)/2 k must be odd. Put k = 2c + 1 u = 2c^2 + 2c + 1 v = (2c + 1)(2c^2 +2c + 1) a = (u + v)/2 = (c + 1)(2c^2 +2c + 1) b = (u - v)/2 = c (2c^2 +2c + 1)
I like the initial substitution a-b=n, but because of the symmetry in the problem I found it easier to continue adding in its brother a+b=m. Then a^2+b^2=(m^2+n^2)/2 and (a-b)^3=n^3. Therefore (m^2+n^2)/2 = n^3. Solve for m^2 to get m^2=n^2*(2n-1). SInce we are working over Z we can conclude 2n-1 is an odd square itself. Let 2n-1=(2c+1)^2. Then n=2c^2+2c+1 and m=+/-(2c+1)*(2c^2+2c+1). For the positive root of m; we have a=(m+n)/2=(c+1)*(2c^2+2c+1) and b=(m-n)/2=c*(2c^2+2c+1) For the negative root of m; we have a=(m+n)/2=-c*(2c^2+2c+1) and b=(m-n)/2=(-c+1)*(2c^2+2c+1) I do think this method is far more elegant as we get the parameterizations without needing to bring in the quadratic formula.
Let a=k*b then (k*b)^2 + b^2 = (k*b -b)^3 assume b not 0 then b= (1+k^2)/(k-1)^3 We must choose k such that b is an integer. Let k=(n+1)/n n an integer. Then b= 1+[(n+1)/n]^2 / 1/(n^3) simplifying b= n*3 + n*(n+1)^2 or b=n*[n^2 +(n+1)^2] a=k*b a=[(n+1)/n]*n*[n^2 +(n+1)^2] = (n+1)*[n^2 +(n+1)^2]. This was one of the formulas derived in the video. Similarly by letting b=k*a you can get the the other formula. I know this method is incomplete because I haven't shown that the substitution k=(n+1)/n is the only value for k that will produce an integer answer for a or b. I don't know how to show that. If someone has a proof please share. But this method does produce a valid parameter method for getting the answers.in the video even though you can't conclude they are all the answers. Just thought this might be interesting.
Nice problem! You can avoid the quadratic by making two substitutions
u = a - b
v = a+ b
We have
(u^2 + v^2)/2 = u^3
Rearranging
v^2 = u^2(2u -1)
Therefore u divides v. Put v = ku
k^2 u^2 = u^2(2u - 1)
k^2 = 2u -1
u = (k^2 + 1)/2
k must be odd. Put k = 2c + 1
u = 2c^2 + 2c + 1
v = (2c + 1)(2c^2 +2c + 1)
a = (u + v)/2 = (c + 1)(2c^2 +2c + 1)
b = (u - v)/2 = c (2c^2 +2c + 1)
Yes, did the same substitutions. At the end we all get the same result but a+b und a-b felt a bit more natural to me...
Isn’t a=0, b=0 a solution as well?
Nice to hear your voice again! I missed it! 🤩
Awsome!
Thank you! 😃
I like the initial substitution a-b=n, but because of the symmetry in the problem I found it easier to continue adding in its brother a+b=m.
Then a^2+b^2=(m^2+n^2)/2 and (a-b)^3=n^3. Therefore (m^2+n^2)/2 = n^3. Solve for m^2 to get m^2=n^2*(2n-1). SInce we are working over Z we can conclude 2n-1 is an odd square itself.
Let 2n-1=(2c+1)^2. Then n=2c^2+2c+1 and m=+/-(2c+1)*(2c^2+2c+1).
For the positive root of m; we have a=(m+n)/2=(c+1)*(2c^2+2c+1) and b=(m-n)/2=c*(2c^2+2c+1)
For the negative root of m; we have a=(m+n)/2=-c*(2c^2+2c+1) and b=(m-n)/2=(-c+1)*(2c^2+2c+1)
I do think this method is far more elegant as we get the parameterizations without needing to bring in the quadratic formula.
Same!
This is an especially nice chapter/video. Thanks!
My pleasure!
Let a=k*b then (k*b)^2 + b^2 = (k*b -b)^3 assume b not 0 then b= (1+k^2)/(k-1)^3 We must choose k such that b is an integer. Let k=(n+1)/n n an integer.
Then b= 1+[(n+1)/n]^2 / 1/(n^3) simplifying b= n*3 + n*(n+1)^2 or b=n*[n^2 +(n+1)^2] a=k*b a=[(n+1)/n]*n*[n^2 +(n+1)^2] = (n+1)*[n^2 +(n+1)^2]. This was one of
the formulas derived in the video. Similarly by letting b=k*a you can get the the other formula. I know this method is incomplete because I haven't shown that the
substitution k=(n+1)/n is the only value for k that will produce an integer answer for a or b. I don't know how to show that. If someone has a proof please share.
But this method does produce a valid parameter method for getting the answers.in the video even though you can't conclude they are all the answers. Just thought
this might be interesting.
Cool - nice to hear your voice again!
Thank you!
How do you get these type of ideas to solve these sums Cyber
a = 0, b = 0
Syber is back!🎉😊
Yay! Thank you!
(a - b)² + 2ab = (a - b)³
[ I ]
a = b =>. a - b = 0
2ab = 0 => *a = b = 0*
[ II ]
a ≠ b
..... I don't know what to do .....
😊🎉😊👍😎
Not bad at all
Thank you! 😍
nice voice came back
had you been cold ?
Change the l to vi 🤪
@@SyberMath really??
terrible ! and you are safe now.good!
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