If a3−b3−3ab=1,\text{ If }a^3 - b^3 - 3ab = 1, If a3−b3−3ab=1,
Prove that (a3−b3+ab) is a perfect square.\text{Prove that }(a^3 - b^3 + ab) \text{ is a perfect square.}Prove that (a3−b3+ab) is a perfect square.
Is the converse true ? \text{Is the converse true ? }Is the converse true ?
Note by Vinay Sipani 7 years, 1 month ago
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
_italics_
**bold**
__bold__
- bulleted- list
1. numbered2. list
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"
\(
\)
\[
\]
2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
The converse is not true, let a=b a = b a=b , we get the second expression a perfect square, while the first is not satisfied.
Now for the proof, we can manipulate the expression hoping to make it simpler , or convert into a quadratic (as it is useful in such situations) we complete the cube to have (a−b)3−1=−3a2b+3ab2+3ab (a-b)^3 -1 = - 3 a^{2} b + 3 a b^{2} + 3ab (a−b)3−1=−3a2b+3ab2+3ab , which is equivalent to (a−b−1)(a2−2ab+b2+a−b+1)=−3ab(a−b−1) (a-b-1)(a^2 - 2ab + b^2 + a - b + 1 ) = -3ab (a-b-1) (a−b−1)(a2−2ab+b2+a−b+1)=−3ab(a−b−1), name this equation (1) so either a=b+1a = b+1a=b+1, in which case a substitution in the the second expression and easy computation show that we have a perfect square (2b+1)2 (2b+1)^2 (2b+1)2, or we have a−b−1≠0 a - b - 1 \neq 0 a−b−1=0, in which case we have by cancelling that factor in (1), the following quadratic equation a2+a(b+1)+b2−b+1=0a^2 + a(b+1) + b^2 - b +1 = 0a2+a(b+1)+b2−b+1=0, but the discriminant of that quadratic equation in aaa is (b+1)2−4(b2−b+1)=−3(b+1)2 (b+1)^2 - 4(b^2-b+1) = -3 ( b +1 )^2 (b+1)2−4(b2−b+1)=−3(b+1)2 , so you have no real solutions except when b=−1 b = -1 b=−1 , for which a=0a = 0 a=0 substituting this in the second expression we get a perfect square.
Log in to reply
ahh thanks @michael_lee sorry, I have made a mistake in the last step it should have been , (b+1)2−4(b2−b+1)=−3(b−1)2 (b+1)^2 -4(b^2-b+1) = -3 (b-1)^2 (b+1)2−4(b2−b+1)=−3(b−1)2 , which shows that the only real solution when a,b a,b a,b are integers we get when b=1 b = 1 b=1 , but at that case a=−1 a = -1 a=−1 , and we don't get a perfect square, so a restatement of proposition must be " If a,ba,b a,b are integers , b≠1 b\neq 1 b=1 and a3−b3−3ab=−1a^3-b^3 -3ab = -1 a3−b3−3ab=−1 , then a3−b3+ab a^3 - b^3 + ab a3−b3+ab is perfect square .
the above user "Mahmoud moustafa" is me , but I have forgotten password and failed to get it back .
Um, neither is true. For example, let a=−1a = -1a=−1, b=1b = 1b=1. Then, a3−b3−3ab=1a^3-b^3-3ab = 1a3−b3−3ab=1 and a3−b3+ab=−3a^3-b^3+ab = -3a3−b3+ab=−3, which is obviously not a perfect square.
In particular, if aaa and bbb are reals such that a≠−1a \neq -1a=−1 or b≠1b \neq 1b=1, then the statement is true. If a3−b3−3ab=1a^3-b^3-3ab = 1a3−b3−3ab=1, then a3−b3+ab=(a3−b3−3ab)+4ab=4ab+1a^3-b^3+ab = (a^3-b^3-3ab)+4ab = 4ab+1a3−b3+ab=(a3−b3−3ab)+4ab=4ab+1. Then, from the first statement, we have a3−3ab−b3−1=(a−b−1)(a2+ab+b2+a−b+1)=0a^3-3ab-b^3-1 = (a-b-1)(a^2+ab+b^2+a-b+1) = 0a3−3ab−b3−1=(a−b−1)(a2+ab+b2+a−b+1)=0. From this, we have a=b+1a = b+1a=b+1, which gives 4ab+1=4(b+1)b+1=4b2+4b+1=(2b+1)24ab+1 = 4(b+1)b+1 = 4b^2+4b+1 = (2b+1)^24ab+1=4(b+1)b+1=4b2+4b+1=(2b+1)2, or a=−b+1±−3b2+6b−32a = -\frac{b+1\pm\sqrt{-3b^2+6b-3}}{2}a=−2b+1±−3b2+6b−3, which is only real for b=1b=1b=1, which gives our solution that does not satisfy the statement, a=−1a = -1a=−1, b=1⇒a3−b3+ab=−3b = 1 \Rightarrow a^3-b^3+ab = -3b=1⇒a3−b3+ab=−3.
Let us assume a-b=1 . ⟹\Longrightarrow ⟹ a3−b3−3ab=a3−b3−3ab(a−b)a^3 -b^3 -3ab=a^3-b^3-3ab(a-b) a3−b3−3ab=a3−b3−3ab(a−b) = (a−b)3 (a-b)^3(a−b)3 . Since a3−b3−3aba^3 -b^3 -3aba3−b3−3ab =1 , (a−b)3 (a-b)^3(a−b)3 is also =1 . Hence a-b=1 . Thus our assumption is valid .Now , a3−b3=(a−b)(a2+ab+b2) a^3-b^3 = (a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2) ⟹ \Longrightarrow \ ⟹ (a3−b3+ab)=(a+b)2(a^{ 3 }-b^{ 3 } +ab) =(a+b)^2 (a3−b3+ab)=(a+b)2 as a−b=1a-b=1 a−b=1 .Hence the above statement is true .
But if (a−b)3=1(a-b)^3 = 1(a−b)3=1 then possible solutions of (a-b) are
(a−b)=1,ω,ω2(a-b)=1,\omega,{\omega}^2(a−b)=1,ω,ω2
How will you justify this ?
No
Problem Loading...
Note Loading...
Set Loading...
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
The converse is not true, let a=b , we get the second expression a perfect square, while the first is not satisfied.
Now for the proof, we can manipulate the expression hoping to make it simpler , or convert into a quadratic (as it is useful in such situations) we complete the cube to have (a−b)3−1=−3a2b+3ab2+3ab , which is equivalent to (a−b−1)(a2−2ab+b2+a−b+1)=−3ab(a−b−1), name this equation (1) so either a=b+1, in which case a substitution in the the second expression and easy computation show that we have a perfect square (2b+1)2, or we have a−b−1=0, in which case we have by cancelling that factor in (1), the following quadratic equation a2+a(b+1)+b2−b+1=0, but the discriminant of that quadratic equation in a is (b+1)2−4(b2−b+1)=−3(b+1)2 , so you have no real solutions except when b=−1 , for which a=0 substituting this in the second expression we get a perfect square.
Log in to reply
ahh thanks @michael_lee sorry, I have made a mistake in the last step it should have been , (b+1)2−4(b2−b+1)=−3(b−1)2 , which shows that the only real solution when a,b are integers we get when b=1 , but at that case a=−1 , and we don't get a perfect square, so a restatement of proposition must be " If a,b are integers , b=1 and a3−b3−3ab=−1 , then a3−b3+ab is perfect square .
the above user "Mahmoud moustafa" is me , but I have forgotten password and failed to get it back .
Um, neither is true. For example, let a=−1, b=1. Then, a3−b3−3ab=1 and a3−b3+ab=−3, which is obviously not a perfect square.
Log in to reply
In particular, if a and b are reals such that a=−1 or b=1, then the statement is true. If a3−b3−3ab=1, then a3−b3+ab=(a3−b3−3ab)+4ab=4ab+1. Then, from the first statement, we have a3−3ab−b3−1=(a−b−1)(a2+ab+b2+a−b+1)=0. From this, we have a=b+1, which gives 4ab+1=4(b+1)b+1=4b2+4b+1=(2b+1)2, or a=−2b+1±−3b2+6b−3, which is only real for b=1, which gives our solution that does not satisfy the statement, a=−1, b=1⇒a3−b3+ab=−3.
Let us assume a-b=1 . ⟹ a3−b3−3ab=a3−b3−3ab(a−b) = (a−b)3 . Since a3−b3−3ab =1 , (a−b)3 is also =1 . Hence a-b=1 . Thus our assumption is valid .Now , a3−b3=(a−b)(a2+ab+b2) ⟹ (a3−b3+ab)=(a+b)2 as a−b=1 .Hence the above statement is true .
Log in to reply
But if (a−b)3=1 then possible solutions of (a-b) are
(a−b)=1,ω,ω2
How will you justify this ?
No