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Question Number 203000 by sonukgindia last updated on 07/Jan/24

Answered by SEKRET last updated on 07/Jan/24

3;6

3;6

Answered by Rasheed.Sindhi last updated on 07/Jan/24

xy^2 +xy+x=129 x(y^2 +y+1)=129=1×3×43 0<x<y ∧ x,y∈N ∧ y>1 y^2 +y+1=3 ∣ y^2 +y+1=43 y^2 +y−2=0 ∣ y^2 +y−42=0 (y+2)(y−1)=0 ∣ (y+7)(y−6)=0 y=1^(×) ,−2^(×) ∣ y=−7^(×) , y=6 ✓ x(y^2 +y+1)=129 ⇒x(6^2 +6+1)=129 ⇒x=((129)/(43))=3 (x,y)=(3,6) 129_(10) =333_6

xy2+xy+x=129x(y2+y+1)=129=1×3×430<x<y∧x,y∈N∧y>1y2+y+1=3∣y2+y+1=43y2+y−2=0∣y2+y−42=0(y+2)(y−1)=0∣(y+7)(y−6)=0y=1×,−2×∣y=−7×,y=6✓x(y2+y+1)=129⇒x(62+6+1)=129⇒x=12943=3(x,y)=(3,6)12910=3336

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