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I reduced a number theory problem to finding all ordered pairs $(a,b)$ that satisfy the equation $a(a-2) = b(b+2)$ in a certain range. After thinking about this for a while, I figured that either $a = b + 2$, $a = -b$, $a = b = 0$ or $a = 2 \text{ and } b = -2$. It is easy to prove that these values for $a$ and $b$ will all satisfy the equation, but how would I solve this equation had I not come up with these solutions? And how do I know I did not miss one? In short, can this equation be solved (more) rigorously?

Edit: Silly of me I did not recognize that my last two solutions were already covered by my first two.

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5 Answers 5

up vote 9 down vote accepted

Add $1$ to both sides to get $$a^2 - 2a + 1 = b^2 + 2b + 1$$ i.e. $$(a-1)^2 = (b+1)^2.$$

Hence $a-1 = b+1$ or $a-1 = -b-1$ and these are all solutions.

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2  
I'd simplify the last expressions to $a=b+2$ or $a=-b$ just to be nice. –  JB King May 30 '13 at 21:27
    
@JBKing Sure, but the OP seemed to have that already. –  mrf May 30 '13 at 21:28

Denote $c=b+2$. Then $(a-1)^2=(c-1)^2$ etc.

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Another approach is to note that $a=b+2$ is an "obvious" solution, then to divide by that solution: $$ \frac{a(a-2)-b(b+2)}{a-(b+2)}=a+b $$ giving $a(a-2)-b(b+2)=(a-(b+2))(a+b)$. Therefore, the solutions are $$ a=b+2\quad\text{and}\quad a=-b $$

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You divided by 0 when you divided by a-b+2, is there a reason why you didn't get an erroneous result from this? –  Ovi May 31 '13 at 0:41
1  
When looking for another solution, $a-(b+2)\ne0$, so it is legal. –  robjohn May 31 '13 at 1:32
    
Oh ok thank you –  Ovi May 31 '13 at 10:08

$a(a - 2) = b(b + 2)$

$a^2 - 2a = b^2 + 2b$

$a^2 - b^2 = 2(b + a)$

$(a + b)(a - b) = 2(a + b)$

$(a + b)(a - b - 2) = 0$

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Put $\, b=x.\, f(x) = x^2\!+\!\color{#c00}2x = a^2\!-\!2a= f(-a)$ has roots $x=-a,\ x'\!=a\!-\!2,$ by $\, x\!+\!x'\!=-\color{#c00}2.$

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