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Give an algebraic proof ?

\[\text{ If }a^3 - b^3 - 3ab = 1,\]

\[\text{Prove that }(a^3 - b^3 + ab) \text{ is a perfect square.}\]

\[\text{Is the converse true ? }\]

Note by Vinay Sipani
3 years ago

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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 \[ (a-b)^3 -1 = - 3 a^{2} b + 3 a b^{2} + 3ab \] , which is equivalent to \[ (a-b-1)(a^2 - 2ab + b^2 + a - b + 1 ) = -3ab (a-b-1) \], 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 \( a - b - 1 \neq 0 \), in which case we have by cancelling that factor in (1), the following quadratic equation \[a^2 + a(b+1) + b^2 - b +1 = 0\], but the discriminant of that quadratic equation in \(a\) is \( (b+1)^2 - 4(b^2-b+1) = -3 ( b +1 )^2 \) , 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. Mahmoud Moustafa · 3 years ago

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@Mahmoud Moustafa ahh thanks @michael_lee sorry, I have made a mistake in the last step it should have been , \( (b+1)^2 -4(b^2-b+1) = -3 (b-1)^2 \) , 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 , \( b\neq 1 \) and \(a^3-b^3 -3ab = -1 \) , then \( a^3 - b^3 + ab \) is perfect square .

the above user "Mahmoud moustafa" is me , but I have forgotten password and failed to get it back . Mahmoud Moustafa · 2 years, 12 months ago

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Um, neither is true. For example, let \(a = -1\), \(b = 1\). Then, \(a^3-b^3-3ab = 1\) and \(a^3-b^3+ab = -3\), which is obviously not a perfect square. Michael Lee · 3 years ago

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@Michael Lee In particular, if \(a\) and \(b\) are reals such that \(a \neq -1\) or \(b \neq 1\), then the statement is true. If \(a^3-b^3-3ab = 1\), then \(a^3-b^3+ab = (a^3-b^3-3ab)+4ab = 4ab+1\). Then, from the first statement, we have \(a^3-3ab-b^3-1 = (a-b-1)(a^2+ab+b^2+a-b+1) = 0\). From this, we have \(a = b+1\), which gives \(4ab+1 = 4(b+1)b+1 = 4b^2+4b+1 = (2b+1)^2\), or \(a = -\frac{b+1\pm\sqrt{-3b^2+6b-3}}{2}\), which is only real for \(b=1\), which gives our solution that does not satisfy the statement, \(a = -1\), \(b = 1 \Rightarrow a^3-b^3+ab = -3\). Michael Lee · 3 years ago

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Let us assume a-b=1 . \(\Longrightarrow \) \(a^3 -b^3 -3ab=a^3-b^3-3ab(a-b) \) = \( (a-b)^3\) . Since \(a^3 -b^3 -3ab\) =1 , \( (a-b)^3\) is also =1 . Hence a-b=1 . Thus our assumption is valid .Now , \( a^3-b^3 = (a-b)(a^2+ab+b^2)\) \( \Longrightarrow \ \) \((a^{ 3 }-b^{ 3 } +ab) =(a+b)^2 \) as \(a-b=1 \) .Hence the above statement is true . Anish Kelkar · 3 years ago

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@Anish Kelkar But if \((a-b)^3 = 1\) then possible solutions of (a-b) are

\((a-b)=1,\omega,{\omega}^2\)

How will you justify this ? Vinay Sipani · 3 years ago

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No Lee Minhyeok · 3 years ago

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