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I am currently practicing my proof writing skills and found a problem that asked to prove that there are not solutions in the natural numbers to :$\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$ ($a$ and $b$ are natural numbers and $a>b$)
I proceeded in the following way:
suppose that: $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\in\mathbb{N}$.
we have :
$\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\in\mathbb{N}$
$\Rightarrow$ $(\exists k\in\mathbb{N}) ;\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=k$
$\Rightarrow$ $a^{2}+b^{2}=k(a^{2}-b^{2})$
$\Rightarrow$ $a^{2}+b^{2}=ka^{2}-kb^{2}$
$\Rightarrow$ $b^{2}(k+1)=a^{2}(k-1)$
$\Rightarrow$ $\frac{b^{2}}{a^{2}}=\frac{k-1}{k+1}$
Notice how $\frac{b^{2}}{a^{2}}$ is positif and smaller than 1, and same for $\frac{k-1}{k+1}$ which contradicts the assumption. Therefore $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\notin\mathbb{N}$

Note: I am not sure of the vality of the reasoning in the last line, and this is the main reason why I am posting to ask a question about it as well: Is treating the case like this is enough or should I analyse the inverse of the last line?

Thank you for anyone who will read this post and thank you for helping ! I wish all of you a wonderful day !

Raid Zougari
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1 Answers1

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When $\frac{b^2}{a^2} = \frac{k-1}{k+1}$, take into account that the difference between both squares, $a^2 - b^2$, in the left side is more than $2$ if $a,b \geq 2$. However, the difference in the right side of the equality, $k + 1 - (k -1) = 2$, which means that there is a contradiction here and that is how we finish showing that $\frac{a^2 + b^2}{a^2-b^2} \notin \mathbb{N}$.
One could have the doubt that what happens if those fractions are simplified. Well, it could happen in the left side depending if $a,b$ share common divisors, but it won't never happen in $\frac{k-1}{k+1}$ since those two numbers cannot have common divisors if $k \geq 2$, which seems the initial case to be considered in this problem. Well, it would be simplified by 2 since they might have the same remainder. I would appreciate if you gave upvote if this answer helped you

Daniel C.
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    Thank you so much for your answer, but could you please clarify why $k-1$ and $k+1$ cannot have a common divisor if $k>=2$. And again thank you so much ! – Raid Zougari Oct 19 '23 at 08:25
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    Well, the common divisor would be only 2. However, having simplified that, they cannot have another since for having a common divisor, they would need to have the same remainding being divided with any $n > 2$. For example, if $(k-1) / 3$ has remainder 2, $(k+1) / 3$ has neccesarily a different remainder. And the same for any other number, 4, 5, and so on. It only works with 2, but this is not a problematic case, it is easy to show. – Daniel C. Oct 19 '23 at 08:28