We can write n in its prime factor form , similarly we can right \(\tau(n)\) as (1+k1)... (1+kr) where k1 ...kr are powers of the prime factors,and then i just tried by establishing simple congruences and relate prime factor and its power but somehow i could not manage that

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TopNewest@Chinmay Sangawadekar- you see divisors come in pairs

if \(d\) divides \(n \) then \(\dfrac{n}{d}\) also divides n

thus you can generate pairs like \((1,n),(2,\dfrac{n}{2}),(3,\dfrac{n}{3})..............(\sqrt{n},\dfrac{n}{\sqrt{n}})\)

thus n can have atmost \(\sqrt n\) pairs or

\(2\sqrt n\) divisors

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I had an idea about the proof somehow using modular arithmetic, lets see if we can come up with fruitful outcomes.

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How to use modular arithmetic here?

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We can write n in its prime factor form , similarly we can right \(\tau(n)\) as (1+k1)... (1+kr) where k1 ...kr are powers of the prime factors,and then i just tried by establishing simple congruences and relate prime factor and its power but somehow i could not manage that

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@Akshat Sharda @Harsh Shrivastava

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