<s>[INST] Why is $ab=(a-b)(a+b)$ false when $a,b$ are coprime? I could not find this, and do not have a background in number theory so sorry if this sounds trivial.
If $a$ and $b$ are coprime integers, why is the statement
$$ab=(a-b)(a+b)$$
a contradiction?
 [/INST] it becomes $$ a^2 - ab - b^2 = 0.   $$ If $b \neq 0,$ divide through by $b^2$ and introduce $r = \frac{a}{b},$ this gives
$$ r^2 - r - 1 = 0  $$ So $r$ is either the Golden ratio
$$ \frac{1 + \sqrt 5}{2} $$ or negative its reciprocal. Put briefly, irrational, contradicting $a,b$ being integers.
It is also possible to use anisotropy in $\mathbb Q_5,$ which is probably what they want. Well, why not. Take
$$ a^2 - ab - b^2 \equiv 0 \pmod {25}   $$  and ASSUME $\gcd(a,b) =1.$
The number $4$ is invertible $\pmod 5 $ and $\pmod {25},$ so we change nothing demanding
$$  4 a^2 - 4ab - 4 b^2  \equiv 0 \pmod {25},  $$
$$  \left( 2a -b \right)^2 - 5 b^2  \equiv 0 \pmod {25}.  $$
It follows that $(2a-b)$ is divisible by $5,$ and then $\left( 2a -b \right)^2$ is divisible by $25.$ In turn, this means $5 b^2$ is divisible by $25,$ so $b^2$ is divisble by $5,$ finally $b$ itself is divisible by $5.$ As $(2a-b)$ is also divisible by $5,$ we find that $a$ is divisible by $5.$ So far, both $a,b$ are divisible by $5.$ This means $5 | \gcd(a,b)$ and $\gcd(a,b) \neq 1$
This is the method I like to show that there is, in fact, no solution of $a^2 - ab - b^2 = 0$ in integers unless both $a,b$ are zero. The lemma that you need to really, really believe, is that, should there be a solution with at least one variable nonzero, then we can divide through by the gcd of those original $a,b$ numbers, divide both by that, resulting in a solution in coprime integers. Our proof of impossibility for coprime integers thus extends to a proof of impossibility  at least one variable nonzero, finally to any nonzero solution. The quadratic form name for this argument is "anisotropy." We say that the quadratic form $a^2 - ab - b^2 $ is not isotropic in the $5$-adic numbers $\mathbb Q_5$
It is also true that the quadratic form $a^2 - ab - b^2 $ is not isotropic in the $2$-adic numbers $\mathbb Q_2.$ This is posted in some earlier answers. One may check, there are just four cases, $a^2 - ab - b^2 $ is odd unless both $a,b$ are even. Thus is was enough to use $\pmod 2,$ for this prime it was not even necessary to use $\pmod 4.$
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