Seeing the "Second solution" (on page 16) to the problem in the post
reminded me of the demonstration, which had long annoyed me, of the fact that $\sqrt{2}$ is irrational. I consulted the textbook when I was in the first year of high school (today's 9th grade) :
BOGDANOF Z., GEORGESCU-BUZAU E., PANAITOPOL L.
- ALGEBRA : Manual pentru anul I licee,
Ed. Did. si Ped., Bucuresti, 1975
Here's what it says on page 23 (starting with the 5th; minor retouching by CiP) :
"There is no rational number whose square is 2.
(in the context of the statement that "the equation $x^2=2$ has no solutions in the set of rational numbers"). The proof is done by contradiction :
Suppose there is a rational number $\frac{m}{n}$ such that $\frac{m^2}{n^2}=2\;\;\;\;\;(1)$
We can assume that the fraction $\frac{m}{n}$ is irreducible (otherwise we simplify with the greatest common divisor of the numbers m and n).
(1)$\;\;\Rightarrow m^2=2n^2\Rightarrow 2\mid m^2\Rightarrow 2\mid m$
so $m=2q\;(q\in\mathbb{Z})$. Then $(2q)^2=2n^2\Leftrightarrow 2q^2=n^2$, so $2\mid n^2$ and then $2\mid n$.
Therefore, the fraction $\frac{m}{n}$ is not irreducible (both the numerator $m$ and the denominator $n$ are divisible by 2). A contradiction was obtained with the hypothesis that the fraction $\frac{m}{n}$ is irreducible."
This made me unhappy, because in fact the contradiction is NOT with the hypothesis, but with an assumption made during the demonstration. It is possible, I have heard, to logically formalize such a path.
(I am also unhappy because I fell in love with a woman ALINA, to whom I plan to dedicate a page on this blog, with our story (I still don't know for sure, at this time, how it will end). She is a big fan of a song.... I also promoted it on Twitter, I'm putting it here too : Three Seconds - Sia Ft Damian Marley
)
Second Solution
If $a=1$ then
$m=\frac{a^2+b}{a+b-1}=\frac{1+b}{b}=1+\frac{1}{b},\;\;n=\frac{b^2+a}{a+b}=\frac{b^2+b-b-1+2}{1+b}=b-1+\frac{2}{1+b}$
and $m,\;n$ are integers if and only if $b=1$.
If $b=1$ then $m=\frac{a^2+1}{a}=a+\frac{1}{a},\;\;n=\frac{1+a}{a+1}=1$ and $m$ is integer if and only if $a=1$.
Let, from now on, $a,\;b \geqslant 2.$
We denote $s=a+b$; in this case we have $s\geqslant 4$.
$n=\frac{a+b+b^2-b}{a+b}=\frac{s+b(b-1)}{s}=1+\frac{b(b-1)}{s} \tag{2}$
$m=\frac{(s-b)^2+b}{s-1}=\frac{s^2-2sb+b^2+b}{s-1}= s-2b+1+\frac{b(b-1)+1}{s-1}\tag{3}$
(For (3) we process the numerator like this $s^2-2sb+b^2+b=(s^2-s)\color{Red}{+s}\underset{=-2b(s-1)}{\underbrace{-2sb+2b}}-b+b^2\color{Red}{-1}+1=$
$=(s-1)(s+1-2b)+b^2-b+1$)
We will show that we can have neither $m\in \mathbb{Z}$ nor $n\in \mathbb{Z}$.
We assume the opposite :
$m\in\mathbb{Z}\;\;AND\;\; n\in\mathbb{Z} \tag{4}$
Because $b(b-1)\geqslant 2$, we have $n\in\mathbb{Z}\;\overset{(2)}{\Rightarrow} \;b(b-1)=k \cdot s,\;k\in\mathbb{N}.$ But $s>b\Rightarrow k\cdot s=b(b-1)>k\cdot b$, hence
$\color{Red}{\fbox{k<b-1}}\tag{5}$
On the other hand $m\in\mathbb{Z}\overset{(3)}{\Rightarrow} s-1\mid ks+1=k(s-1)+(k+1)$, so $s-1 \mid k+1$, and hence $k+1\geqslant s-1$ from where
$k\geqslant s-2=a+b-2=(a-1)+(b-1)$.
From the previous row it follows :
$\color{Red}{\fbox{$k\geqslant b-1$}}\tag{6}$
But (5) and (6) are contradictory.
And here, as in the first example, the contradiction appeared somewhere you didn't expect it. So (4) is false, so at least one of the numbers $m,\;n$ is not an integer.
