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Question Number 53144 by mr W last updated on 18/Jan/19
Find all integers x and y such that  ((xy)/(x+y)) is also integer.
$${Find}\:{all}\:{integers}\:{x}\:{and}\:{y}\:{such}\:{that} \\ $$$$\frac{{xy}}{{x}+{y}}\:{is}\:{also}\:{integer}. \\ $$
Commented by mr W last updated on 19/Jan/19
x=(i+1)j  y=i(i+1)j  i,j ∈ Z
$$\boldsymbol{{x}}=\left(\boldsymbol{{i}}+\mathrm{1}\right)\boldsymbol{{j}} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{i}}\left(\boldsymbol{{i}}+\mathrm{1}\right)\boldsymbol{{j}} \\ $$$$\boldsymbol{{i}},\boldsymbol{{j}}\:\in\:\mathbb{Z} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19
   x=0  y=1  ((xy)/(x+y))=0  and x=1 y=0  ((xy)/(x+y))=0    ((xy)/(x+y)) is integer when  1)  x+y=1 x=1,2,3,4...then y=0,−1,−2,−3...  2)x+y=1  y=1,2,3,4...then x=0,−1,−2 ...  sir i am trying to find more...  x+y=−1  x=0,1 ,2,3...y=−1,−2,−3...  x+y=−1  y=0,1,2,3...x=−1,−2,−3,...
$$\:\:\:{x}=\mathrm{0}\:\:{y}=\mathrm{1}\:\:\frac{{xy}}{{x}+{y}}=\mathrm{0}\:\:{and}\:{x}=\mathrm{1}\:{y}=\mathrm{0}\:\:\frac{{xy}}{{x}+{y}}=\mathrm{0} \\ $$$$ \\ $$$$\frac{{xy}}{{x}+{y}}\:{is}\:{integer}\:{when} \\ $$$$\left.\mathrm{1}\right)\:\:{x}+{y}=\mathrm{1}\:{x}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}…{then}\:{y}=\mathrm{0},−\mathrm{1},−\mathrm{2},−\mathrm{3}… \\ $$$$\left.\mathrm{2}\right){x}+{y}=\mathrm{1}\:\:{y}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}…{then}\:{x}=\mathrm{0},−\mathrm{1},−\mathrm{2}\:… \\ $$$${sir}\:{i}\:{am}\:{trying}\:{to}\:{find}\:{more}… \\ $$$${x}+{y}=−\mathrm{1}\:\:{x}=\mathrm{0},\mathrm{1}\:,\mathrm{2},\mathrm{3}…{y}=−\mathrm{1},−\mathrm{2},−\mathrm{3}… \\ $$$${x}+{y}=−\mathrm{1}\:\:{y}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}…{x}=−\mathrm{1},−\mathrm{2},−\mathrm{3},… \\ $$
Commented by mr W last updated on 18/Jan/19
thanks for trying!  can you give a general solution for  all solutions?
$${thanks}\:{for}\:{trying}! \\ $$$${can}\:{you}\:{give}\:{a}\:{general}\:{solution}\:{for} \\ $$$${all}\:{solutions}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19
sir i am trying...
$${sir}\:{i}\:{am}\:{trying}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19
i think general solution related to greatest integer  function...better you post genetal solution sir...        and
$${i}\:{think}\:{general}\:{solution}\:{related}\:{to}\:{greatest}\:{integer} \\ $$$${function}…{better}\:{you}\:{post}\:{genetal}\:{solution}\:{sir}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${and}\: \\ $$
Answered by ajfour last updated on 18/Jan/19
(1/N)=(1/x)+(1/y) = ((x+y)/(xy))  ⇒   z^2 −z+N = 0  x,y = ((1±(√(1−4N)))/2)  ⇒  1−4N = (2k−1)^2   ⇒  −4N = 4k^2 −4k  or  N = k−k^2   x, y = ((1±∣2k−1∣)/2) .
$$\frac{\mathrm{1}}{{N}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:=\:\frac{{x}+{y}}{{xy}} \\ $$$$\Rightarrow\:\:\:{z}^{\mathrm{2}} −{z}+{N}\:=\:\mathrm{0} \\ $$$${x},{y}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}{N}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{1}−\mathrm{4}{N}\:=\:\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:−\mathrm{4}{N}\:=\:\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{k} \\ $$$${or}\:\:{N}\:=\:{k}−{k}^{\mathrm{2}} \\ $$$${x},\:{y}\:=\:\frac{\mathrm{1}\pm\mid\mathrm{2}{k}−\mathrm{1}\mid}{\mathrm{2}}\:.\:\:\:\:\:\:\: \\ $$
Commented by ajfour last updated on 18/Jan/19
will this do Sir ?
$${will}\:{this}\:{do}\:{Sir}\:? \\ $$
Commented by mr W last updated on 18/Jan/19
it describs only some solutions, i.e.  those with x+y=1  but not all solutions, so it is not  a general solution sir, e.g.  following solutions are correct, but they  can not be obtained from your formula:  x=2, y=2  x=3, y=6
$${it}\:{describs}\:{only}\:{some}\:{solutions},\:{i}.{e}. \\ $$$${those}\:{with}\:{x}+{y}=\mathrm{1} \\ $$$${but}\:{not}\:{all}\:{solutions},\:{so}\:{it}\:{is}\:{not} \\ $$$${a}\:{general}\:{solution}\:{sir},\:{e}.{g}. \\ $$$${following}\:{solutions}\:{are}\:{correct},\:{but}\:{they} \\ $$$${can}\:{not}\:{be}\:{obtained}\:{from}\:{your}\:{formula}: \\ $$$${x}=\mathrm{2},\:{y}=\mathrm{2} \\ $$$${x}=\mathrm{3},\:{y}=\mathrm{6} \\ $$
Commented by ajfour last updated on 18/Jan/19
okay sir, i suspected so too, shall  try again.
$${okay}\:{sir},\:{i}\:{suspected}\:{so}\:{too},\:{shall} \\ $$$${try}\:{again}. \\ $$

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