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This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.

The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.

So I started by expanding $(a−b)^2$ to $(a−b)^2 = (a-b)(a-b) = a^2 -2ab +b^2$. To Prove that $(a−b)^2 = a^2 −b^2 $ if b = 0 I substituted b with zero both in the expanded expression and the original simplified and I got $(a−b)^2 = (a-0)^2 = (a-0)(a-0) = a^2 - a(0)-a(0)+0^2 = a^2$ and the same with $a^2 -2ab +b^2$ which resulted in $a^2 - 2a(0) + 0^2 = 2a$ or if I do not substite the $b^2$ I end up with $a^2 + b^2$. That's what I got when I try to prove the expression true for $b=0$.

As for the part where $b=a$, $(a−b)^2 = (a-b)(a-b) = a^2-2ab+b^2$, if a and b are equal, let $a=b=x$ and I substite $a^2-2ab+b^2 = x^2-2(x)(x) + x^2 = x^2-2x^2+x^2 = 1-2+1=0$ I do not see where any of this can be reduced to $a^2-b^2$ unless that equals zero......I do see where it holds but I do not see how would a solution writting out look.After typing this it seems a lot clearer but I just can't see how to phrase a "solution".

P.S: This is my first time asking a question here so whatever I did wrong I am sorry in advance and appreciate the feedback.

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    You are only proving b = 0 or b=a $\implies(a-b)^2 = a^2 - b^2$. You also have to prove $(a-b)^2 = a^2 - b^2 \implies $ b= 0 or b = a. – 2017-01-24
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    First of all! I want to thank everyone for their prompt responses and effort to help. I reached a possible solution that I would appreciate feedback on: First of all, $(a-b)^2 = (a-b)(a-b) = a^2 -2ab +b^2$ and on the other hand: $a^2 -b^2 = (a+b)(a-b)$ which leaves us with $(a-b)(a-b) = (a+b)(a-b)$ and the only way for that expression to be true is if first, $b=0$, then $(a-0)(a-0) = (a+0)(a-0) = a$ or if $a=b$ then $(a-b)(a-b) = (a+b)(a-b) = (0)(0) = (a+b)(0) = 0 – 2017-01-24
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    btw should I add my above comment to the question or should I post it as an answer to my own question? – 2017-01-24

6 Answers 6

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Hint: $(a−b)^2 = a^2 −b^2 \iff (a−b)^2 - a^2 +b^2=0 \iff 2b(b-a)=0\,$.

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It might just be easier to use that $a^2-b^2=(a-b)(a+b)$.

So if $a-b=0$ then $(a-b)^2=(a-b)(a+b)$, and if $a-b\neq 0$ then $(a-b)^2=(a-b)(a+b)$ if and only if $a-b=a+b$.

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    That was how I saw it, too. – 2017-01-24
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which resulted in $a^2 - 2a(0) + 0^2 = 2a$

$a^2 - 2a(0) + 0^2 = a^2$, not $2a$.

or if I do not substitute the $b^2$ I end up with $a^2 + b^2$.

Why would you not substitute the $b^2$? If you're substituting $b=0$ then you need to do it in all occurrences of $b$. This includes $b^2$. When you do this you'll see that the equation $(a-b)^2 = a^2 - b^2$ reduces to $a^2 = a^2$, which is certainly a true statement for all values of $a$.

Also, you're overcomplicating the $a=b$ case. No need to introduce a new variable $x$. If $a = b$, then you can simply substitute either one in for the other. Let's replace $a$ with $b$. Then we have $$ (a-b)^2 = (b-b)^2 = 0^2 = 0$$ and on the other hand we have $$ a^2 - b^2 = b^2 - b^2 = 0 $$ This shows that $(a-b)^2 = a^2 - b^2$ if $a = b$.

I should also point out that the work you've done (or at least the work you've shown us) only proves one direction. You're asked to prove the following: $$ (a-b)^2 = a^2 - b^2 \text{ if and only if } b = 0 \text{ or } a = b $$ But what you've done so far is: $$ (a-b)^2 = a^2 - b^2 \text{ if } b = 0 \text{ or } a = b $$

In other words, you need to handle the "only if" part. Do this by assuming $(a-b)^2 = a^2 - b^2$ and showing that the only two possibilities are $b=0$ or $a=b$. At least one of the other current answers offers guidance on this matter.

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This is a geometrical approach.

Let $a-b=d$. Then $d^2 + b^2 = a^2$ and the using Converse of Pythagoras theorem there must be a triangle having $a, b, a-b$ as sides, which is impossible because $b + (a - b) = a \not \gt a$.

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we have $$(a-b)^2=a^2-b^2$$ we know that $$(a-b)^2=a^2+b^2-2ab$$ and if we use the first equation we get $$2b^2-2ab=0$$ thus we get $$b=0$$ or $$b=a$$

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    why can we not use $$(a-b)^2=a^2-2ab+b^2$$ ? – 2017-01-24
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    Nobody said you can't. It would help if you tried to solve the problem asked, however. – 2017-01-24
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    i have solved the problem if you wold read my post – 2017-01-24
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    The original equation involves $a^2-b^2$ not $a^2+b^2$ so you don't get to $ab=0$ – 2017-01-24
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    it is corrected sorry – 2017-01-24
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You have to prove two things:

1) If $b = 0$ or $b = a$ then $(a-b)^2 = a^2 - b^2$.

And

2) If $(a-b)^2 = a^2 - b^2$ then either $b = 0 $ or $b = a$.

To prove 1: we do what you did correctly:

If $b = 0$ then $(a - b)^2 = (a-0)^2 = a^2$. An $a^2 - b^2 = a^2 - 0^2 = a^2 - 0 = a^2$. So $(a - b)^2= a^2 - b^2$.

If $b =a$ then $(a- b)^2 = (a-a)^2 = 0^2 = 0$. and $a^2 - b^2 = a^2 - b^2 = 0$. SO $(a- b)^2 = a^2 - b^2$.

That was the easy part.

To prove 2)

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

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

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

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

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

$-2ab + 2b^2 = 0$

$2b(a- b) = 0$

Now if $mn = 0$ either $m = 0$ or $n=0$; that should be a basic fact you know.

So either $2b = 0$ or $a-b = 0$.

If $2b = 0$ then $b = 0$. If $a-b = 0$ then $a = b$.

So if $(a-b)^2 = a^2 -b^2$ then either $b =0$ or $b= a$.