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Question Number 27767 by Rasheed.Sindhi last updated on 14/Jan/18
If the number of divisors of a number is odd,prove that the number is perfect square and vice versa.
$$\mathrm{If}\:\mathrm{the}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{divisors}}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{number}\:\mathrm{is}\:\boldsymbol{\mathrm{odd}},\mathrm{prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{is}\:\boldsymbol{\mathrm{perfect}}\:\boldsymbol{\mathrm{square}}\:\mathrm{and} \\ $$$$\mathrm{vice}\:\mathrm{versa}. \\ $$
Answered by mrW2 last updated on 14/Jan/18
Every number n can be expressed as n=Π_(i=1) ^k p_i ^e_i , where p_i is prime, and the number of its divisors is d(n)=Π_(i=1) ^k (e_i +1) If a number N is a perfect square, i.e. N=n^2 , then N=(Π_(i=1) ^k p_i ^e_i )^2 =Π_(i=1) ^k p_i ^(2e_i ) d(N)=Π_(i=1) ^k (2e_i +1) since the product of odd numbers is odd, therefore d(N) is odd. If d(n)=Π_(i=1) ^k (e_i +1) is odd, it means e_i must be even, let′s say e_i =2f_i , then n=Π_(i=1) ^k p_i ^e_i =Π_(i=1) ^k p_i ^(2f_i ) =(Π_(i=1) ^k p_i ^f_i )^2 it means n is a perfect square.
$${Every}\:{number}\:{n}\:{can}\:{be}\:{expressed}\:{as} \\ $$$${n}=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{{e}_{{i}} } ,\:{where}\:{p}_{{i}} \:{is}\:{prime}, \\ $$$${and}\:{the}\:{number}\:{of}\:{its}\:{divisors}\:{is} \\ $$$${d}\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\left({e}_{{i}} +\mathrm{1}\right) \\ $$$$ \\ $$$${If}\:{a}\:{number}\:{N}\:{is}\:{a}\:{perfect}\:{square}, \\ $$$${i}.{e}.\:{N}={n}^{\mathrm{2}} ,\:{then} \\ $$$${N}=\left(\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{{e}_{{i}} } \right)^{\mathrm{2}} =\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{\mathrm{2}{e}_{{i}} } \\ $$$${d}\left({N}\right)=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\left(\mathrm{2}{e}_{{i}} +\mathrm{1}\right) \\ $$$${since}\:{the}\:{product}\:{of}\:{odd}\:{numbers}\:{is}\:{odd}, \\ $$$${therefore}\:{d}\left({N}\right)\:{is}\:{odd}. \\ $$$$ \\ $$$${If}\:{d}\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\left({e}_{{i}} +\mathrm{1}\right)\:{is}\:{odd},\:{it}\:{means}\:{e}_{{i}} \\ $$$${must}\:{be}\:{even},\:{let}'{s}\:{say}\:{e}_{{i}} =\mathrm{2}{f}_{{i}} ,\:{then} \\ $$$${n}=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{{e}_{{i}} } =\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{\mathrm{2}{f}_{{i}} } =\left(\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{p}_{{i}} ^{{f}_{{i}} } \right)^{\mathrm{2}} \\ $$$${it}\:{means}\:{n}\:{is}\:{a}\:{perfect}\:{square}. \\ $$
Commented by mrW2 last updated on 15/Jan/18
An other way to prove: Generally, if m is a divisor of the number N, then (N/m) is also a divisor of it. That means the divisors are always pairwise. Every divisor has a corresponding partner. If the number N is perfect square, i.e. N=n^2 , it means n is a divisor of it, but (N/n)=n, the partner of n is in fact itself, it means if the number is perfect square, its number of divisors is odd. On the other side, if the number of divisors of N is odd, one divisor, say n, and its partner has the same value, i.e. n=(N/n) ⇒N=n^2 or N is perfect square.
$${An}\:{other}\:{way}\:{to}\:{prove}: \\ $$$${Generally},\:{if}\:{m}\:{is}\:{a}\:{divisor}\:{of}\:{the}\: \\ $$$${number}\:{N},\:{then}\:\frac{{N}}{{m}}\:{is}\:{also}\:{a}\:{divisor} \\ $$$${of}\:{it}.\:{That}\:{means}\:{the}\:{divisors}\:{are} \\ $$$${always}\:{pairwise}.\:{Every}\:{divisor}\:{has} \\ $$$${a}\:{corresponding}\:{partner}. \\ $$$$ \\ $$$${If}\:{the}\:{number}\:{N}\:{is}\:{perfect}\:{square},\:{i}.{e}. \\ $$$${N}={n}^{\mathrm{2}} ,\:{it}\:{means}\:{n}\:{is}\:{a}\:{divisor}\:{of}\:{it}, \\ $$$${but}\:\frac{{N}}{{n}}={n},\:{the}\:{partner}\:{of}\:{n}\:{is}\:{in}\:{fact} \\ $$$${itself},\:{it}\:{means} \\ $$$${if}\:{the}\:{number}\:{is}\:{perfect}\:{square},\:{its} \\ $$$${number}\:{of}\:{divisors}\:{is}\:{odd}. \\ $$$$ \\ $$$${On}\:{the}\:{other}\:{side},\:{if}\:{the}\:{number}\:{of} \\ $$$${divisors}\:{of}\:{N}\:{is}\:{odd},\:{one}\:{divisor},\:{say}\:\:{n}, \\ $$$${and}\:{its}\:{partner}\:{has}\:{the}\:{same}\:{value},\:{i}.{e}. \\ $$$${n}=\frac{{N}}{{n}} \\ $$$$\Rightarrow{N}={n}^{\mathrm{2}} \\ $$$${or}\:{N}\:{is}\:{perfect}\:{square}. \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jan/18
Thanks Very Much Sir! Knowledge increasing answer for me!
$$\:\mathcal{T}{hanks}\:{Very}\:{Much}\:{Sir}! \\ $$$${Knowledge}\:{increasing}\:{answer}\:{for}\:{me}! \\ $$

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