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September 19th, 2018, 08:11 AM   #11
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(1 - 2m)^3 = (-8) m^3 + 12 m^2 - 6 m + 1 = 3 n^2 + 3 n + 1
how to find n?

Last edited by skipjack; September 19th, 2018 at 10:31 AM.
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September 19th, 2018, 10:32 AM   #12
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The right-hand side is the difference between the cubes of consecutive integers; as this difference is a perfect cube, it must be 1, so n is -1 or 0, and m = 0.
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September 19th, 2018, 03:01 PM   #13
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( b +1)^3-b^3=The right-hand side

but the difference between consecutive cubes
is not a cube.
1^3+(3b^2+3b+1)=3b^2+3b+2
1+ 3*1+3.1+1 = 3 +3 +2
1+ 7 = 8
=2^3

n=-1 et 0
m=0 is it a complete study on this subject?

Last edited by skipjack; September 19th, 2018 at 08:15 PM.
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September 19th, 2018, 08:26 PM   #14
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Quote:
Originally Posted by Ak23 View Post
. . . but the difference between consecutive cubes
is not a cube.
The equation states that its right-hand side, 3 n^2 + 3 n + 1, which is the difference between n³ and (n + 1)³, equals its left-hand side, which is (1 - 2m)³. Consecutive cubes differ by a perfect cube only when they are -1 and 0 or 0 and 1.
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September 20th, 2018, 12:56 AM   #15
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thank you very much
I bring here the support of 2n + 1 = x ^ 2?
if x is odd we can write it 1-2m
x ^ 2 = 4m ^ 2 - 4m + 1
= 2 (2m ^ 2 - 2m) + 1
= 2n + 1
and
4m ^ 2 - 4m + 1 = 2n + 1
(4m ^ 2 - 4m) = 2n
(4m ^ 2 - 4m) / 2 = n
2m ^ 2 - 2m = n
2 (m ^ 2 - m) = n


From what you showed me.
Do we get this with (1-2m)^3 if not show how to do it?

Last edited by skipjack; September 20th, 2018 at 03:05 AM. Reason: to correct typos and sign error
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September 20th, 2018, 02:34 AM   #16
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If x^2 = 2n + 1 = (1 - 2m)^2
sqrt(2n + 1) = |1 - 2m|
n = ((-2m + 1)^2 - 1)/2
for any m $\in$ N

Last edited by skipjack; September 20th, 2018 at 02:54 AM. Reason: to correct
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September 20th, 2018, 03:00 AM   #17
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(1 - 2m)³ = (1 - 2m)(1 - 2m)² = (1 - 2m)(1 - 4m + 4m²) = 1 - 6m + 12m² - 8m³

In your last post, I made some corrections.

It would make more sense to say "for any m $\small\in$ ℤ" than "for any m $\small\in$ N".
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September 20th, 2018, 03:43 AM   #18
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1-2m=(3n^2+3n+1)^1/3
there are no solutions because there is the negative root
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September 20th, 2018, 03:53 AM   #19
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Quote:
It would make more sense to say "for any m $\small\in$ ℤ" than "for any m $\small\in$ N".
I did not understand
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September 20th, 2018, 04:34 AM   #20
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There's no reason to exclude negative values of m.

Quote:
Originally Posted by Ak23 View Post
. . . there are no solutions because there is the negative root
There is no negative cube root of 3n² + 3n + 1.
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