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Question Number 216995 by Rasheed.Sindhi last updated on 26/Feb/25

Find all prime numbers  p and q   such that  p^2 −  q^2 =  2024

Findallprimenumberspandqsuchthatp2q2=2024

Answered by Marzuk last updated on 26/Feb/25

  (p+q)(p−q)=2024  2024 = 2×1012             = 4×506             =8× 253             =88×23             =1×2024             =22×92             =11×184             =44×46  p+q+p−q=2p  that means sum of the each pair has  to be divisible by 2  2+1012=1014  4+506=510  8+253=261  88+23=111  1+2024=2025  22+92=114  11+184=195  44+46=90  so we have left 4 pairs  2×1012  4×506  22×92  44×46  lets see if the dividend is the sumof each  pair and the divisor is 2 then the quotient  is ∈ P  ((Σ(2,1012))/2)  ∉ P  ((Σ(4,506))/2) ∉ P  ((Σ(22,92))/2) ∉ P  ((Σ(44,46))/2) ∉ P   So,the set of all possible prime values of  is ∅

(p+q)(pq)=20242024=2×1012=4×506=8×253=88×23=1×2024=22×92=11×184=44×46p+q+pq=2pthatmeanssumoftheeachpairhastobedivisibleby22+1012=10144+506=5108+253=26188+23=1111+2024=202522+92=11411+184=19544+46=90sowehaveleft4pairs2×10124×50622×9244×46letsseeifthedividendisthesumofeachpairandthedivisoris2thenthequotientisPΣ(2,1012)2PΣ(4,506)2PΣ(22,92)2PΣ(44,46)2PSo,thesetofallpossibleprimevaluesofis

Commented by Rasheed.Sindhi last updated on 26/Feb/25

Thanks sir!

Thankssir!

Answered by mehdee7396 last updated on 26/Feb/25

2024=1012×2=506×4=46×44=92×22  1)p+q=1012  & p−q=2⇒p=507  & q=505  ×  2)p+q=506  & p−q=4⇒p=255  & q=251  ×  3)p+q=46  & p−q=44⇒p=45  & q=1  ×  4)p+q=92  & p−q=22⇒p=57  & q=35  ×  there is no answer

2024=1012×2=506×4=46×44=92×221)p+q=1012&pq=2p=507&q=505×2)p+q=506&pq=4p=255&q=251×3)p+q=46&pq=44p=45&q=1×4)p+q=92&pq=22p=57&q=35×thereisnoanswer

Commented by Rasheed.Sindhi last updated on 26/Feb/25

Thanx sir!

Thanxsir!

Answered by MathematicalUser2357 last updated on 01/Mar/25

can′t find all prime numbers p and q because I tried to find it!

cantfindallprimenumberspandqbecauseItriedtofindit!

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