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what-is-the-number-of-ordered-pairs-of-positif-integers-x-y-that-satisfy-x-2-y-2-xy-37-




Question Number 100178 by bobhans last updated on 25/Jun/20
what is the number of ordered pairs of positif   integers (x,y) that satisfy x^2 +y^2 −xy=37
whatisthenumberoforderedpairsofpositifintegers(x,y)thatsatisfyx2+y2xy=37
Answered by 1549442205 last updated on 25/Jun/20
x^2 −xy+y^2 −37=0.We look at it like a  quadratic equation respect with x.Then  Δ=y^2 −4y^2 +148=−3y^2 +148.The above  eqs.always has roots ,so Δ=−3y^2 +148≥0  ⇒y^2 ≤((148)/3).Since y∈N^(∗ ) ,y∈{1,2,...,7}.  i)If y∈{1,2,5,6}then Δ∈{145,136,73,40}  aren′t perfect numbers,so are rejected  ii)for y=3⇒Δ=121=11^2 ⇒x=((3+11)/2)=7(we reject negative root)  we get the root (x,y)=(7;3)  iii)for y=4 ⇒Δ=100=10^2 ⇒x=((4+10)/2)=7  we get the root (x;y)=(7;4)  iv)for y=7 ⇒Δ=1⇒x=((7±1)/2)⇒x∈{4;3}  we get two roots (x;y)∈{(4;7);(3;7)}  Thus ,the given equation has four roots  (x;y)∈{(7;3);(7;4);(3;7);(4;7)}
x2xy+y237=0.Welookatitlikeaquadraticequationrespectwithx.ThenΔ=y24y2+148=3y2+148.Theaboveeqs.alwayshasroots,soΔ=3y2+1480y21483.SinceyN,y{1,2,,7}.i)Ify{1,2,5,6}thenΔ{145,136,73,40}arentperfectnumbers,soarerejectedii)fory=3Δ=121=112x=3+112=7(werejectnegativeroot)wegettheroot(x,y)=(7;3)iii)fory=4Δ=100=102x=4+102=7wegettheroot(x;y)=(7;4)iv)fory=7Δ=1x=7±12x{4;3}wegettworoots(x;y){(4;7);(3;7)}Thus,thegivenequationhasfourroots(x;y){(7;3);(7;4);(3;7);(4;7)}
Commented by bemath last updated on 25/Jun/20
how with (4,−3),(−4,3),(3,−4),(−3,4)?
howwith(4,3),(4,3),(3,4),(3,4)?
Commented by 1549442205 last updated on 25/Jun/20
your problem  require to find the pairs of   positive integers
yourproblemrequiretofindthepairsofpositiveintegers
Answered by Rasheed.Sindhi last updated on 25/Jun/20
A Novel Method_(−)   x^2 +y^2 −xy=37  (x−y)^2 =37−xy............A  (x+y)^2 =37+3xy............B  x,y∈Z^+  ∧ 37−xy is perfect square  37−xy=36,25,16,9,4,1(possible values)  xy=1,12,21,28,33,36  From above values of xy acceptable  values are those for which 37+3xy  is also perfect square are 21 & 28  When xy=21  A⇒x−y=4 (assume x≥y)  B⇒x+y=10  x=7,y=3        (or x=3,y=7 due to symmetry)  When xy=28  A⇒x−y=3  B⇒x+y=11        x=7,y=4   (or x=4,y=7 due to symmetry)  (7,3),(3,7),(7,4),(4,7)
ANovelMethodx2+y2xy=37(xy)2=37xyA(x+y)2=37+3xyBx,yZ+37xyisperfectsquare37xy=36,25,16,9,4,1(possiblevalues)xy=1,12,21,28,33,36Fromabovevaluesofxyacceptablevaluesarethoseforwhich37+3xyisalsoperfectsquareare21&28Whenxy=21Axy=4(assumexy)Bx+y=10x=7,y=3(orx=3,y=7duetosymmetry)Whenxy=28Axy=3Bx+y=11x=7,y=4(orx=4,y=7duetosymmetry)(7,3),(3,7),(7,4),(4,7)
Commented by mr W last updated on 25/Jun/20
nice solution sir!
nicesolutionsir!
Commented by bobhans last updated on 26/Jun/20
thank you sir
thankyousir

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