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Question-211454




Question Number 211454 by BaliramKumar last updated on 09/Sep/24
Answered by A5T last updated on 09/Sep/24
Let N=p_1 ^2 ×p_2 ^4 ;then it has (2+1)(4+1)=15 divisors  ⇒N^2 =p_1 ^4 ×p_2 ^8 ; then it has (4+1)(8+1)=45
$${Let}\:{N}={p}_{\mathrm{1}} ^{\mathrm{2}} ×{p}_{\mathrm{2}} ^{\mathrm{4}} ;{then}\:{it}\:{has}\:\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{4}+\mathrm{1}\right)=\mathrm{15}\:{divisors} \\ $$$$\Rightarrow{N}^{\mathrm{2}} ={p}_{\mathrm{1}} ^{\mathrm{4}} ×{p}_{\mathrm{2}} ^{\mathrm{8}} ;\:{then}\:{it}\:{has}\:\left(\mathrm{4}+\mathrm{1}\right)\left(\mathrm{8}+\mathrm{1}\right)=\mathrm{45} \\ $$
Answered by mr W last updated on 11/Sep/24
case 1: N=p^a   a+1=15 ⇒a=14  p≥2 ⇒N≥2^(14) =16384 >9999   ⇒no solution    case 2: N=p^a q^b   (a+1)(b+1)=15=3×5 ⇒a, b=2, 4  2^2 5^4 =2500 ✓  2^2 7^4 =9604 ✓  3^2 5^4 =5625 ✓ 5^2 3^4 =2025 ✓  3^4 7^2 =3969 ✓  ⇒there are 5 solutions  N^2 =p^(2a) q^(2b)  has (2a+1)(2b+1)  =5×9=45 divisors
$${case}\:\mathrm{1}:\:{N}={p}^{{a}} \\ $$$${a}+\mathrm{1}=\mathrm{15}\:\Rightarrow{a}=\mathrm{14} \\ $$$${p}\geqslant\mathrm{2}\:\Rightarrow{N}\geqslant\mathrm{2}^{\mathrm{14}} =\mathrm{16384}\:>\mathrm{9999}\: \\ $$$$\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{N}={p}^{{a}} {q}^{{b}} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)=\mathrm{15}=\mathrm{3}×\mathrm{5}\:\Rightarrow{a},\:{b}=\mathrm{2},\:\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{4}} =\mathrm{2500}\:\checkmark \\ $$$$\mathrm{2}^{\mathrm{2}} \mathrm{7}^{\mathrm{4}} =\mathrm{9604}\:\checkmark \\ $$$$\mathrm{3}^{\mathrm{2}} \mathrm{5}^{\mathrm{4}} =\mathrm{5625}\:\checkmark\:\mathrm{5}^{\mathrm{2}} \mathrm{3}^{\mathrm{4}} =\mathrm{2025}\:\checkmark \\ $$$$\mathrm{3}^{\mathrm{4}} \mathrm{7}^{\mathrm{2}} =\mathrm{3969}\:\checkmark \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{5}\:{solutions} \\ $$$${N}^{\mathrm{2}} ={p}^{\mathrm{2}{a}} {q}^{\mathrm{2}{b}} \:{has}\:\left(\mathrm{2}{a}+\mathrm{1}\right)\left(\mathrm{2}{b}+\mathrm{1}\right) \\ $$$$=\mathrm{5}×\mathrm{9}=\mathrm{45}\:{divisors} \\ $$

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