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Question Number 212515 by efronzo1 last updated on 16/Oct/24
Thenumbersofpairsofnaturalnumbers(x,y)withx,y⩽33thatsatisfy5∣3xy−1+2y2xis...(A)295(B)296(C)297(D)298(E)299
Commented by A5T last updated on 16/Oct/24
I′mgetting313pairs,doyouhaveanysolutiontothis?
Answered by A5T last updated on 16/Oct/24
3xy−1+2y2x≡(−2)xy−1+2y2x≡0(mod5)Notethatwhenx(=2k)isevenandy⩾3(−2)xy−1+2y2x=((−2)2y−1)ky−1+2y4k≡0(mod5)⇒2y4k≡4(mod5)⇒y4k−2≡0(mod4)⇒y4k≡2(mod4)→←So,x(=2k+1)isoddwheny⩾3wheny=1;3xy−1+2y2x≡3+2≡0(mod5)foranyx⇒(x,y)=(1,1),(2,1),(3,1),...,(33,1)⇒33solutionswheny=2;3xy−1+2y2x≡3x+24x≡3x+1≡0(mod5)⇒3x≡4(mod5)⇒y=2∧x=4q+2⇒8solutionswheny⩾3,weknowthatxisoddwhenx=1;3xy−1+2y2x≡3+2y2≡0(mod5)⇒2y2≡2(mod5)⇒y2≡1(mod4)⇒y=4l+1or4l+3⩾3⇒(x,y)=(1,3),(1,5),...,(1,33)⇒16solutionsSupposey(=2m+1)⩾3isalsooddandx⩾33xy−1+2y2x=3(2k+1)2m+2(2m+1)4k+2⇒2(2m+1)4k+2≡2(2k+1)2m(mod5)⇒(2m+1)4k+2≡(2k+1)2m(mod4)⇒1+(2m)1(4k+2)≡1+(2k)(2m)(mod4)whichisalwaystrueforallmandk(⩾1)⇒16×16=256solutionsWheny(=2m)iseven.⇒3(2k+1)2m−1+2(2m)4k+2≡0(mod5)⇒(2m)4k+2≡(2k+1)2m−1(mod4)→←⇒Totalnumberofsolutions=256+16+8+33=313
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