Let $a,b\in\mathbb{N}$.
Prove that if $1+ab\equiv 0[24]$ then $a+b\equiv 0[24]$
My attempts:
$1+ab\equiv 0[24]\implies ab\equiv -1[24]$
so : $ab\equiv 23[24]$
Then there $k\in\mathbb {N}$ such as $ab=24k$
But how can I move to $a+b$?.
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Bill Dubuque
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Ellen Ellen
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1Having arrived at the fact that $ab\equiv 23\mod(24)$, your next question should be: what are the factors of $23$ modulo $24$? List them in pairs, then add them modulo $24$. What do you get? – H. sapiens rex Mar 26 '24 at 09:56