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Question Number 50369 by prof Abdo imad last updated on 16/Dec/18
find x ,y from Z  wich verify  y^2 =x(x+1)(x+7)(x+8)
findx,yfromZwichverifyy2=x(x+1)(x+7)(x+8)
Answered by mr W last updated on 16/Dec/18
let t=x+4  y^2 =(t−4)(t−3)(t+3)(t+4)  y^2 =(t^2 −16)(t^2 −9)  let s=t^2   y^2 =(s−16)(s−9)  s^2 −25s+144−y^2 =0  s=((25±(√(25^2 −4(144−y^2 ))))/2)  s=((25±(√(4y^2 +49)))/2)  4y^2 +49=(2n+1)^2   49=(2n+1)^2 −(2y)^2 =(2n+1−2y)(2n+1+2y)  2n+1−2y=1⇒n=y  2n+1+2y=49⇒4y=48⇒y=12  with y=12:  ⇒s=25 or 0  ⇒t=5 or 0  ⇒x=1 or −4    solutions are:  x=−4, y=12  x=1, y=12
lett=x+4y2=(t4)(t3)(t+3)(t+4)y2=(t216)(t29)lets=t2y2=(s16)(s9)s225s+144y2=0s=25±2524(144y2)2s=25±4y2+4924y2+49=(2n+1)249=(2n+1)2(2y)2=(2n+12y)(2n+1+2y)2n+12y=1n=y2n+1+2y=494y=48y=12withy=12:s=25or0t=5or0x=1or4solutionsare:x=4,y=12x=1,y=12
Commented by maxmathsup by imad last updated on 16/Dec/18
thank you sir for this hardwork.
thankyousirforthishardwork.
Commented by Necxx last updated on 17/Dec/18
please why did you equate 4y^2 +49=(2n+1)^2 ???
pleasewhydidyouequate4y2+49=(2n+1)2???
Commented by mr W last updated on 17/Dec/18
s=((25±(√(4y^2 +49)))/2)  such that s is integer,  (√(4y^2 +49)) must be an odd integer, say  2n+1, therefore  4y^2 +49=(2n+1)^2
s=25±4y2+492suchthatsisinteger,4y2+49mustbeanoddinteger,say2n+1,therefore4y2+49=(2n+1)2

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