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If I have a triad of natural numbers $a$, $b$ and $c$, and another triad $a'$, $b'$ and $c'$, such that either $a\neq a'$, $b\neq b'$, or $c\neq c'$, is it possible that these equations are true?

$ab+c=a'b'+c'$

$ac+b=a'c'+b'$

$cb+a=c'b'+a'$

If I had only the first equation, it is obviously possible, for example $a=8$, $b=2$ and $c=14$ give the same result as $a'=10$, $b'=2$ and $c'=10$, when we introduce the other two equations it feels like it would be impossible but I don't know how to prove it

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There are (perhaps some might say trivial) counterexamples of the form $(1,m,n)$ and $(1,n,m)$--for instance $a=1,b=3,c=4$ and $a'=1, b'=4, c'=3$

There are also nontrivial counterexamples such as: $$a=2,b=5,c=89 {\rm \;\; and\;\; } a'=5,b'=13, c'=34$$

I found this with a search using Python.

I just noticed: They are all Fibonacci numbers.

Here's one that isn't all Fibonacci numbers: $$a=2,b=9,c=67 {\rm \;\; and\;\; } a'=3,b'=14, c'=43$$

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  • $\begingroup$ Thank you very much! The relationship with Fibonacci was very surprising, do you have any ideas of why it shows up? $\endgroup$
    – Andrés Romero
    Jun 30, 2022 at 0:52
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    $\begingroup$ I have no idea why the one with all Fibonacci number's showed up! $\endgroup$
    – paw88789
    Jun 30, 2022 at 0:53
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    $\begingroup$ It's very interesting, thank you again! $\endgroup$
    – Andrés Romero
    Jun 30, 2022 at 1:02

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