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Question Number 135507 by 0731619177 last updated on 13/Mar/21
Answered by Rasheed.Sindhi last updated on 15/Mar/21
(x+y)(81+xy) − 9(x+y)^2 =210 (x+y)(81+xy−9(x+y))=210 Let x+y=n where n∈Z ∧ n ∣ 210 x+y=n ∧ 81+xy−9(x+y)=210/n y=n−x→ 81+x(n−x)−9n=210/n xn^2 −x^2 n−3(3n^2 −27n+70)=0 x^2 n−xn^2 +3(3n^2 −27n+70)=0 ( Quadratic in x) Discrimenent (D): D=n^4 −12n(3n^2 −27n+70) As x is an integer so D must be perfect square. possible values for n: ±1,±2,±3,±5,±6,±7,±10,±14,±15, ±21,±30,±35,±70,±105,±210 Only for n=5,6,7; D is perfect square: D=25,144,49 So only n=5,6,7 may be valid values. ^• (x+y)(81+xy−9(x+y))=5×42 x+y=5 ∧ 81+xy−9(x+y)=42 81+xy−9(5)=42 xy=6 x=2,y=3 or x=3,y=2 ^• (x+y)(81+xy−9(x+y))=6×35 x+y=6 ∧ 81+xy−9(x+y)=35 x+y=6 ∧ xy=8 x=2,y=4 or x=4,y=2 ^• (x+y)(81+xy−9(x+y))=7×30 x+y=7 ∧ 81+xy−9(x+y)=30 x+y=7 ∧ xy=12 x=3,y=4 or x=4,y=3 (x,y)=(2,3),(3,2),(2,4),(4,2),(3,4),(4,3)
$$\left({x}+{y}\right)\left(\mathrm{81}+{xy}\right)\:−\:\mathrm{9}\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{210} \\ $$$$\left({x}+{y}\right)\left(\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)\right)=\mathrm{210} \\ $$$$ \\ $$$${Let}\:{x}+{y}={n}\:{where}\:{n}\in\mathbb{Z}\:\wedge\:{n}\:\mid\:\mathrm{210} \\ $$$${x}+{y}={n}\:\wedge\:\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)=\mathrm{210}/{n} \\ $$$${y}={n}−{x}\rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{81}+{x}\left({n}−{x}\right)−\mathrm{9}{n}=\mathrm{210}/{n} \\ $$$$\:\:\:\:\:\:\:\:{xn}^{\mathrm{2}} −{x}^{\mathrm{2}} {n}−\mathrm{3}\left(\mathrm{3}{n}^{\mathrm{2}} −\mathrm{27}{n}+\mathrm{70}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} {n}−{xn}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{3}{n}^{\mathrm{2}} −\mathrm{27}{n}+\mathrm{70}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left(\:{Quadratic}\:{in}\:{x}\right) \\ $$$${Discrimenent}\:\left({D}\right): \\ $$$$\:\:\:\:{D}={n}^{\mathrm{4}} −\mathrm{12}{n}\left(\mathrm{3}{n}^{\mathrm{2}} −\mathrm{27}{n}+\mathrm{70}\right) \\ $$$${As}\:{x}\:{is}\:{an}\:{integer}\:{so}\:{D}\:{must}\:{be}\:{perfect} \\ $$$${square}. \\ $$$${possible}\:{values}\:{for}\:{n}: \\ $$$$\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3},\pm\mathrm{5},\pm\mathrm{6},\pm\mathrm{7},\pm\mathrm{10},\pm\mathrm{14},\pm\mathrm{15}, \\ $$$$\pm\mathrm{21},\pm\mathrm{30},\pm\mathrm{35},\pm\mathrm{70},\pm\mathrm{105},\pm\mathrm{210} \\ $$$${Only}\:{for}\:{n}=\mathrm{5},\mathrm{6},\mathrm{7};\:{D}\:\:{is}\:{perfect}\:{square}: \\ $$$$\:{D}=\mathrm{25},\mathrm{144},\mathrm{49} \\ $$$${So}\:{only}\:{n}=\mathrm{5},\mathrm{6},\mathrm{7}\:\boldsymbol{{may}}\:\boldsymbol{{be}}\:{valid}\:{values}. \\ $$$$\:^{\bullet} \left({x}+{y}\right)\left(\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)\right)=\mathrm{5}×\mathrm{42} \\ $$$$\:\:\:{x}+{y}=\mathrm{5}\:\wedge\:\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)=\mathrm{42} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{81}+{xy}−\mathrm{9}\left(\mathrm{5}\right)=\mathrm{42} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xy}=\mathrm{6} \\ $$$${x}=\mathrm{2},{y}=\mathrm{3}\:{or}\:{x}=\mathrm{3},{y}=\mathrm{2} \\ $$$$\:^{\bullet} \left({x}+{y}\right)\left(\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)\right)=\mathrm{6}×\mathrm{35} \\ $$$$\:\:\:\:{x}+{y}=\mathrm{6}\:\wedge\:\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)=\mathrm{35} \\ $$$$\:\:\:\:\:{x}+{y}=\mathrm{6}\:\wedge\:{xy}=\mathrm{8} \\ $$$$\:\:\:\:{x}=\mathrm{2},{y}=\mathrm{4}\:{or}\:{x}=\mathrm{4},{y}=\mathrm{2} \\ $$$$\:^{\bullet} \left({x}+{y}\right)\left(\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)\right)=\mathrm{7}×\mathrm{30} \\ $$$$\:\:\:\:{x}+{y}=\mathrm{7}\:\wedge\:\mathrm{81}+{xy}−\mathrm{9}\left({x}+{y}\right)=\mathrm{30} \\ $$$$\:\:\:{x}+{y}=\mathrm{7}\:\wedge\:{xy}=\mathrm{12} \\ $$$$\:\:\:{x}=\mathrm{3},{y}=\mathrm{4}\:{or}\:{x}=\mathrm{4},{y}=\mathrm{3} \\ $$$$ \\ $$$$\left({x},{y}\right)=\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{3},\mathrm{2}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{4},\mathrm{2}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right) \\ $$
Commented by mr W last updated on 19/Mar/21
very nice solution sir!
$${very}\:{nice}\:{solution}\:{sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 19/Mar/21
THαnKs Sir! Actually I felt that the answer is unnoticed at all! Even the questioner isn′t given any type of response!! So I request you to see it! Thanks for response.
$$\mathcal{TH}\alpha{n}\mathcal{K}{s}\:\mathcal{S}{ir}! \\ $$$$\mathcal{A}{ctually}\:{I}\:{felt}\:{that}\:{the}\:{answer}\:{is} \\ $$$${unnoticed}\:{at}\:{all}!\:\mathcal{E}{ven}\:{the}\:{questioner} \\ $$$${isn}'{t}\:{given}\:{any}\:{type}\:{of}\:{response}!! \\ $$$$\mathcal{S}{o}\:{I}\:{request}\:{you}\:{to}\:{see}\:{it}! \\ $$$${Thanks}\:{for}\:{response}. \\ $$
Commented by mr W last updated on 19/Mar/21
maybe he is just not used to give feedback at all, as some other people. i tried in a little different way and got exactly the same results as you.
$${maybe}\:{he}\:{is}\:{just}\:{not}\:{used}\:{to}\:{give} \\ $$$${feedback}\:{at}\:{all},\:{as}\:{some}\:{other}\:{people}. \\ $$$${i}\:{tried}\:{in}\:{a}\:{little}\:{different}\:{way}\:{and} \\ $$$${got}\:{exactly}\:{the}\:{same}\:{results}\:{as}\:{you}. \\ $$
Commented by Rasheed.Sindhi last updated on 20/Mar/21
Please S^(H^A R) E your way S^I R! I′m interested in an alternate.
$$\mathcal{P}{lease}\:\:\mathcal{S}^{\mathcal{H}^{\mathcal{A}} \mathcal{R}} \mathcal{E}\:\:{your}\:{way}\:\:\mathcal{S}^{\mathcal{I}} \mathcal{R}! \\ $$$${I}'{m}\:{interested}\:{in}\:{an}\:{alternate}. \\ $$
Answered by mr W last updated on 20/Mar/21
9(x+y)^2 −(81+xy)(x+y)+210=0 Δ=(81+xy)^2 −7560=n^2 ⇒(xy+81nn)(xy+81−n)=7560=pq ⇒xy+81+n=p ∧ xy+81−n=q ⇒xy=((p+q)/2)−81^(∗∗)) ∧ n=((p−q)/2)^(∗)) ^∗ ) p≥q, p and q are both even or odd ^(∗∗) ) ((p+q)/2)>81 ⇒p+q>162 x+y=((xy+81±n)/(18))=((p or q)/(18)) ⇒ at least p or q is a multiple of 18. pq=7560 ⇒q<(√(7560))=87 if q=18m<87 ⇒m<5 pq=18×2^2 ×3×5×7 only following cases canfulfill above: (with p underlined) pq =18×2×210 ⇒x+y=2∧xy=42 × =18×3×140 ⇒x+y=3∧xy=16 × =18×5×84 ⇒x+y=5∧xy=6 ✓ =18×6×70 ⇒x+y=6∧xy=8 ✓ =18×7×60 ⇒x+y=7∧xy=12 ✓ =18×10×42 ⇒x+y=10∧xy=30 × no need to go further, since xy should ≤(((x+y)/2))^2 , but upon here xy>(((x+y)/2))^2 . summary: we got following solutions x,y=(2,3), (2,4), (3,4)
$$\mathrm{9}\left({x}+{y}\right)^{\mathrm{2}} −\left(\mathrm{81}+{xy}\right)\left({x}+{y}\right)+\mathrm{210}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{81}+{xy}\right)^{\mathrm{2}} −\mathrm{7560}={n}^{\mathrm{2}} \\ $$$$\Rightarrow\left({xy}+\mathrm{81}{nn}\right)\left({xy}+\mathrm{81}−{n}\right)=\mathrm{7560}={pq} \\ $$$$\Rightarrow{xy}+\mathrm{81}+{n}={p}\:\wedge\:{xy}+\mathrm{81}−{n}={q} \\ $$$$\Rightarrow{xy}=\frac{{p}+{q}}{\mathrm{2}}−\mathrm{81}\:^{\left.\ast\ast\right)} \:\wedge\:{n}=\frac{{p}−{q}}{\mathrm{2}}\:^{\left.\ast\right)} \\ $$$$\left.\:^{\ast} \right)\:{p}\geqslant{q},\:{p}\:{and}\:{q}\:{are}\:{both}\:{even}\:{or}\:{odd} \\ $$$$\left.\:^{\ast\ast} \right)\:\frac{{p}+{q}}{\mathrm{2}}>\mathrm{81}\:\Rightarrow{p}+{q}>\mathrm{162} \\ $$$${x}+{y}=\frac{{xy}+\mathrm{81}\pm{n}}{\mathrm{18}}=\frac{{p}\:{or}\:{q}}{\mathrm{18}}\: \\ $$$$\Rightarrow\:{at}\:{least}\:{p}\:{or}\:{q}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{18}. \\ $$$$ \\ $$$${pq}=\mathrm{7560}\:\Rightarrow{q}<\sqrt{\mathrm{7560}}=\mathrm{87} \\ $$$${if}\:{q}=\mathrm{18}{m}<\mathrm{87}\:\Rightarrow{m}<\mathrm{5} \\ $$$$ \\ $$$${pq}=\mathrm{18}×\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$${only}\:{following}\:{cases}\:{canfulfill}\:{above}: \\ $$$$\left({with}\:{p}\:{underlined}\right) \\ $$$${pq}\:=\mathrm{18}×\mathrm{2}×\underline{\mathrm{210}}\:\Rightarrow{x}+{y}=\mathrm{2}\wedge{xy}=\mathrm{42}\:× \\ $$$$\:\:\:\:\:\:=\mathrm{18}×\mathrm{3}×\underline{\mathrm{140}}\:\Rightarrow{x}+{y}=\mathrm{3}\wedge{xy}=\mathrm{16}\:× \\ $$$$\:\:\:\:\:\:=\underline{\mathrm{18}×\mathrm{5}}×\mathrm{84}\:\Rightarrow{x}+{y}=\mathrm{5}\wedge{xy}=\mathrm{6}\:\checkmark \\ $$$$\:\:\:\:\:\:=\underline{\mathrm{18}×\mathrm{6}}×\mathrm{70}\:\Rightarrow{x}+{y}=\mathrm{6}\wedge{xy}=\mathrm{8}\:\checkmark \\ $$$$\:\:\:\:\:\:=\underline{\mathrm{18}×\mathrm{7}}×\mathrm{60}\:\Rightarrow{x}+{y}=\mathrm{7}\wedge{xy}=\mathrm{12}\:\checkmark \\ $$$$\:\:\:\:\:\:=\underline{\mathrm{18}×\mathrm{10}}×\mathrm{42}\:\Rightarrow{x}+{y}=\mathrm{10}\wedge{xy}=\mathrm{30}\:× \\ $$$$\:\:\:\:\:\:{no}\:{need}\:{to}\:{go}\:{further},\:{since}\: \\ $$$$\:\:\:\:\:\:{xy}\:{should}\:\leqslant\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} ,\:{but}\:{upon} \\ $$$$\:\:\:\:\:\:{here}\:{xy}>\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} . \\ $$$$ \\ $$$${summary}:\:{we}\:{got}\:{following}\:{solutions} \\ $$$${x},{y}=\left(\mathrm{2},\mathrm{3}\right),\:\left(\mathrm{2},\mathrm{4}\right),\:\left(\mathrm{3},\mathrm{4}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 20/Mar/21
Better Sir!
$$\mathcal{B}{etter}\:\mathcal{S}{ir}! \\ $$
Commented by mr W last updated on 20/Mar/21
thanks sir! our methods are similar.
$${thanks}\:{sir}!\:{our}\:{methods}\:{are}\:{similar}. \\ $$
Commented by Rasheed.Sindhi last updated on 21/Mar/21
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