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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
2 years, 8 months 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. · 2 years, 8 months ago

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 . · 2 years, 7 months ago

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. · 2 years, 7 months ago

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, $$a = -1$$, $$b = 1 \Rightarrow a^3-b^3+ab = -3$$. · 2 years, 7 months ago

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 . · 2 years, 7 months ago

But if $$(a-b)^3 = 1$$ then possible solutions of (a-b) are

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

How will you justify this ? · 2 years, 7 months ago

No · 2 years, 8 months ago

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