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Question Number 113689 by Ar Brandon last updated on 14/Sep/20
Find p and q such that p^2 +q^2 =101^2 . Where p, q∈Z different from zero.
$$\mathrm{Find}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{101}^{\mathrm{2}} .\:\mathrm{Where}\:\mathrm{p},\:\mathrm{q}\in\mathbb{Z}\: \\ $$$$\mathrm{different}\:\mathrm{from}\:\mathrm{zero}. \\ $$
Commented by Rasheed.Sindhi last updated on 14/Sep/20
p=20,q=99 or p=99,q=20
$${p}=\mathrm{20},{q}=\mathrm{99}\:{or}\:{p}=\mathrm{99},{q}=\mathrm{20} \\ $$
Answered by nimnim last updated on 14/Sep/20
If a,b and c are primitive pythagorean triples then:− Case(1) when a=odd, b=(a^2 /2)−(1/2) and c=(a^2 /2)+(1/2) Let a=13, c=((169)/2)+(1/2)=85≠101 a=15, c=((225)/2)+(1/2)=113≠101 Case(2) when a=even b=((a/2))^2 −1 and c=((a/2))^2 +1 Let a=20, c=(((20)/2))^2 +1=101 b=(((20)/2))^2 −1=99 ⇒(p,q)=(20,99) or (99,20)
$$\mathrm{If}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{are}\:\mathrm{primitive}\:\mathrm{pythagorean}\:\mathrm{triples} \\ $$$$\mathrm{then}:− \\ $$$$\mathrm{Case}\left(\mathrm{1}\right)\:\mathrm{when}\:\mathrm{a}=\mathrm{odd}, \\ $$$$\:\:\:\:\:\:\:\:\mathrm{b}=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{c}=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{Let}\:\mathrm{a}=\mathrm{13},\:\mathrm{c}=\frac{\mathrm{169}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{85}\neq\mathrm{101} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{a}=\mathrm{15},\:\mathrm{c}=\frac{\mathrm{225}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{113}\neq\mathrm{101} \\ $$$$\:\mathrm{Case}\left(\mathrm{2}\right)\:\mathrm{when}\:\mathrm{a}=\mathrm{even} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}=\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}\:\mathrm{and}\:\mathrm{c}=\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\:\mathrm{Let}\:\mathrm{a}=\mathrm{20},\:\mathrm{c}=\left(\frac{\mathrm{20}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{101} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{b}=\left(\frac{\mathrm{20}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{99} \\ $$$$\Rightarrow\left(\mathrm{p},\mathrm{q}\right)=\left(\mathrm{20},\mathrm{99}\right)\:\mathrm{or}\:\left(\mathrm{99},\mathrm{20}\right) \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 15/Sep/20
If m & n are any positive integers with m>n (m^2 −n^2 ,2mn,m^2 +n^2 ) is a pathagorean triplet. This rule is perfect,it can produce all the pathagorean triplets and that also means that for every triplet there exist m & n which can produce that triplet. Here (p,q,101) is pathagorean triplet so m^2 +n^2 =101 or m^2 =101−n^2 . Clearly n≤⌊(√(101)) ⌋,so we′ve to find such n for which 101−n^2 is perfect square between 1 and 10. Hence n=1,10 are only proper values and corresponding values of m are 10 ,1.Since m>n so (m,n)=(10,1) p=m^2 −n^2 =10^2 −1^2 =99 q=2mn=2(10)(1)=20 Or p=2mn=20 q=m^2 −n^2 =99 Since p,q∈Z (p,q)=(20,99),(20,−99), (−20,99),(−20,−99), (99,20),(99,−20), (−99,20),(−99,−20).
$${If}\:{m}\:\&\:{n}\:\:{are}\:\:\boldsymbol{{any}}\:{positive} \\ $$$${integers}\:{with}\:{m}>{n} \\ $$$$\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} ,\mathrm{2}{mn},{m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)\:{is}\:{a} \\ $$$${pathagorean}\:{triplet}. \\ $$$$\mathcal{T}{his}\:{rule}\:{is}\:{perfect},{it}\:{can}\:{produce} \\ $$$$\boldsymbol{{all}}\:{the}\:{pathagorean}\:{triplets}\:{and} \\ $$$${that}\:{also}\:{means}\:{that}\:{for}\:\boldsymbol{{every}} \\ $$$${triplet}\:{there}\:{exist}\:{m}\:\&\:{n}\:{which} \\ $$$${can}\:{produce}\:{that}\:{triplet}. \\ $$$${Here}\:\left(\mathrm{p},\mathrm{q},\mathrm{101}\right)\:{is}\:{pathagorean} \\ $$$${triplet}\:{so}\:\:\:{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{101} \\ $$$${or}\:{m}^{\mathrm{2}} =\mathrm{101}−{n}^{\mathrm{2}} .\:{Clearly} \\ $$$${n}\leqslant\lfloor\sqrt{\mathrm{101}}\:\rfloor,{so}\:{we}'{ve}\:{to}\:{find}\:{such} \\ $$$${n}\:{for}\:{which}\:\mathrm{101}−{n}^{\mathrm{2}} \:{is}\:{perfect} \\ $$$${square}\:{between}\:\mathrm{1}\:{and}\:\mathrm{10}. \\ $$$${Hence}\:{n}=\mathrm{1},\mathrm{10}\:{are}\:{only}\:{proper}\: \\ $$$${values}\:{and}\:{corresponding}\:{values} \\ $$$${of}\:{m}\:{are}\:\mathrm{10}\:,\mathrm{1}.{Since}\:{m}>{n}\:{so} \\ $$$$\left({m},{n}\right)=\left(\mathrm{10},\mathrm{1}\right) \\ $$$${p}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} =\mathrm{99} \\ $$$${q}=\mathrm{2}{mn}=\mathrm{2}\left(\mathrm{10}\right)\left(\mathrm{1}\right)=\mathrm{20} \\ $$$${Or} \\ $$$${p}=\mathrm{2}{mn}=\mathrm{20} \\ $$$${q}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} =\mathrm{99} \\ $$$${Since}\:{p},{q}\in\mathbb{Z} \\ $$$$\left({p},{q}\right)=\left(\mathrm{20},\mathrm{99}\right),\left(\mathrm{20},−\mathrm{99}\right), \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{20},\mathrm{99}\right),\left(−\mathrm{20},−\mathrm{99}\right), \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{99},\mathrm{20}\right),\left(\mathrm{99},−\mathrm{20}\right), \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{99},\mathrm{20}\right),\left(−\mathrm{99},−\mathrm{20}\right). \\ $$
Answered by 1549442205PVT last updated on 16/Sep/20
We known that( a,b,c )is a triple of Pythagor posirive integer numbers where b^2 +c^2 =a^2 (a,b)=1if and only if c=uv,b=((u^2 −v^2 )/2), a=((u^2 +v^2 )/2) where u,v are odd and coprime .In the our problem p^2 +q^2 =101^2 (∗) Since 101 is a prime,so (p,q)=1. Hence,p=uv,q=((u^2 −v^2 )/2),((u^2 +v^2 )/2)=101 ⇔u^2 +v^2 =202⇒v<9<u≤13 i)If u=13 then v^2 =202−169=33⇒rejected ii)If u=11 then v^2 =202−121=81=9^2 ⇒v=9⇒p=uv=99,q=((11^2 −9^2 )/2)=20 Since p,q have equal role in the equation,we obtain (p,q)∈{(99,20),(20,99).From this follows (−p,q),(p,−q),(−p,−q) also satisfying second way: If(p,q)is a root of (∗) then (−p,q),(p,−q) and (−p,−q)are also roots of (∗).Hence we just need find pairs (p,q)of positive integer numbers of the equation(∗) Since 101 is a prime number,p and q are coprime( because suppose that gcd(p,q)=d≠1 then p=md,q=nd with (m,n)=1.Then p^2 +q^2 =101^2 (∗) ⇔(m^2 +n^2 )d^2 =101^2 ⇒101⋮d which is contradiction to the fact that 101 is prime) .Furthermore,among two numbers p,q must have one number is even and other number is odd.Hence,WLOG suppose p is odd and q is even.From that we have (∗)⇔p^2 =(101+q)(101−q) Put 101+q=m,101−q=n ⇒m>n,we get p^2 =mn,q=((m−n)/2),101=((m+n)/2). We see that m and n −coprime. because of suppose the contrary, that gcd(m,n)=d≠1.Then m=m_1 d,n=n_1 d From that 101=((m+n)/2)=((m_1 +n_1 )/2)d which is impossible since 101 is prime Therefore ,m and n are coprime.From p^2 =mn and (m,n)=1 we infer m=u^2 ,n=v^2 where u and v−coprime, u>v >0 and are odd numbers.Finally, we obtain { ((p=uv(1))),((q=((u^2 −v^2 )/2)(2))),((101=((u^2 +v^2 )/2)(3))) :} (3)⇔u^2 +v^2 =202⇒v<11≤u≤13 i)For u=13⇒v^2 =202−169=33⇒∄v∈Z ii)For u=11⇒v^2 =202−121=81 ⇒v=9.Put into (1)(2) we get (p,q)∈(99,20)and due to above note and the equality of p and q in equation we get (p,q)∈{99,20),(−99,20),(20,99),(20,−99), (99,−20),(−99,−20),(−20,99),(−20,−99)}
$$\mathrm{We}\:\mathrm{known}\:\mathrm{that}\left(\:\mathrm{a},\mathrm{b},\mathrm{c}\:\right)\mathrm{is}\:\mathrm{a}\:\mathrm{triple}\:\mathrm{of}\: \\ $$$$\mathrm{Pythagor}\:\mathrm{posirive}\:\mathrm{integer}\:\mathrm{numbers}\: \\ $$$$\mathrm{where}\:\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)=\mathrm{1if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{c}=\mathrm{uv},\mathrm{b}=\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}, \\ $$$$\mathrm{a}=\frac{\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{where}\:\mathrm{u},\mathrm{v}\:\mathrm{are}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{coprime} \\ $$$$.\mathrm{In}\:\mathrm{the}\:\mathrm{our}\:\mathrm{problem}\:\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{101}^{\mathrm{2}} \left(\ast\right) \\ $$$$\mathrm{Since}\:\mathrm{101}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime},\mathrm{so}\:\left(\mathrm{p},\mathrm{q}\right)=\mathrm{1}. \\ $$$$\mathrm{Hence},\mathrm{p}=\mathrm{uv},\mathrm{q}=\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} }{\mathrm{2}},\frac{\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{101} \\ $$$$\Leftrightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{202}\Rightarrow\mathrm{v}<\mathrm{9}<\mathrm{u}\leqslant\mathrm{13} \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{u}=\mathrm{13}\:\mathrm{then}\:\mathrm{v}^{\mathrm{2}} =\mathrm{202}−\mathrm{169}=\mathrm{33}\Rightarrow\mathrm{rejected} \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{u}=\mathrm{11}\:\mathrm{then}\:\mathrm{v}^{\mathrm{2}} =\mathrm{202}−\mathrm{121}=\mathrm{81}=\mathrm{9}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{v}=\mathrm{9}\Rightarrow\mathrm{p}=\mathrm{uv}=\mathrm{99},\mathrm{q}=\frac{\mathrm{11}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{20} \\ $$$$\mathrm{Since}\:\mathrm{p},\mathrm{q}\:\:\mathrm{have}\:\:\mathrm{equal}\:\mathrm{role}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{equation},\mathrm{we}\:\mathrm{obtain}\: \\ $$$$\left(\boldsymbol{\mathrm{p}},\boldsymbol{\mathrm{q}}\right)\in\left\{\left(\mathrm{99},\mathrm{20}\right),\left(\mathrm{20},\mathrm{99}\right).\mathrm{From}\:\mathrm{this}\right. \\ $$$$\mathrm{follows}\:\left(−\mathrm{p},\mathrm{q}\right),\left(\mathrm{p},−\mathrm{q}\right),\left(−\mathrm{p},−\mathrm{q}\right) \\ $$$$\mathrm{also}\:\mathrm{satisfying} \\ $$$$\boldsymbol{\mathrm{second}}\:\boldsymbol{\mathrm{way}}: \\ $$$$\mathrm{If}\left(\mathrm{p},\mathrm{q}\right)\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\left(\ast\right)\:\mathrm{then}\:\left(−\mathrm{p},\mathrm{q}\right),\left(\mathrm{p},−\mathrm{q}\right) \\ $$$$\mathrm{and}\:\left(−\mathrm{p},−\mathrm{q}\right)\mathrm{are}\:\mathrm{also}\:\mathrm{roots}\:\mathrm{of}\:\left(\ast\right).\mathrm{Hence} \\ $$$$\mathrm{we}\:\mathrm{just}\:\mathrm{need}\:\mathrm{find}\:\mathrm{pairs}\:\left(\mathrm{p},\mathrm{q}\right)\mathrm{of}\:\mathrm{positive} \\ $$$$\mathrm{integer}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\left(\ast\right) \\ $$$$\mathrm{Since}\:\mathrm{101}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number},\mathrm{p}\:\mathrm{and}\:\mathrm{q} \\ $$$$\mathrm{are}\:\mathrm{coprime}\left(\:\mathrm{because}\:\mathrm{suppose}\:\mathrm{that}\:\right. \\ $$$$\mathrm{gcd}\left(\mathrm{p},\mathrm{q}\right)=\mathrm{d}\neq\mathrm{1}\:\mathrm{then}\:\mathrm{p}=\mathrm{md},\mathrm{q}=\mathrm{nd}\:\mathrm{with} \\ $$$$\left(\mathrm{m},\mathrm{n}\right)=\mathrm{1}.\mathrm{Then}\:\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{101}^{\mathrm{2}} \left(\ast\right) \\ $$$$\Leftrightarrow\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)\mathrm{d}^{\mathrm{2}} =\mathrm{101}^{\mathrm{2}} \Rightarrow\mathrm{101}\vdots\mathrm{d}\:\mathrm{which}\:\mathrm{is} \\ $$$$\left.\mathrm{contradiction}\:\mathrm{to}\:\mathrm{the}\:\mathrm{fact}\:\mathrm{that}\:\mathrm{101}\:\mathrm{is}\:\mathrm{prime}\right) \\ $$$$.\mathrm{Furthermore},\mathrm{among}\:\mathrm{two}\:\mathrm{numbers}\:\mathrm{p},\mathrm{q} \\ $$$$\mathrm{must}\:\mathrm{have}\:\mathrm{one}\:\mathrm{number}\:\mathrm{is}\:\mathrm{even}\:\mathrm{and}\:\mathrm{other} \\ $$$$\mathrm{number}\:\mathrm{is}\:\mathrm{odd}.\mathrm{Hence},\mathrm{WLOG}\:\mathrm{suppose} \\ $$$$\mathrm{p}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{q}\:\mathrm{is}\:\mathrm{even}.\mathrm{From}\:\mathrm{that}\:\mathrm{we}\:\mathrm{have} \\ $$$$\:\left(\ast\right)\Leftrightarrow\mathrm{p}^{\mathrm{2}} =\left(\mathrm{101}+\mathrm{q}\right)\left(\mathrm{101}−\mathrm{q}\right) \\ $$$$\mathrm{Put}\:\mathrm{101}+\mathrm{q}=\mathrm{m},\mathrm{101}−\mathrm{q}=\mathrm{n}\:\Rightarrow\mathrm{m}>\mathrm{n},\mathrm{we} \\ $$$$\mathrm{get}\:\mathrm{p}^{\mathrm{2}} =\mathrm{mn},\mathrm{q}=\frac{\mathrm{m}−\mathrm{n}}{\mathrm{2}},\mathrm{101}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{2}}. \\ $$$$\mathrm{We}\:\mathrm{see}\:\mathrm{that}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:−\mathrm{coprime}. \\ $$$$\mathrm{because}\:\mathrm{of}\:\mathrm{suppose}\:\mathrm{the}\:\mathrm{contrary},\:\mathrm{that} \\ $$$$\mathrm{gcd}\left(\mathrm{m},\mathrm{n}\right)=\mathrm{d}\neq\mathrm{1}.\mathrm{Then}\:\mathrm{m}=\mathrm{m}_{\mathrm{1}} \mathrm{d},\mathrm{n}=\mathrm{n}_{\mathrm{1}} \mathrm{d} \\ $$$$\mathrm{From}\:\mathrm{that}\:\mathrm{101}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{2}}=\frac{\mathrm{m}_{\mathrm{1}} +\mathrm{n}_{\mathrm{1}} }{\mathrm{2}}\mathrm{d} \\ $$$$\mathrm{which}\:\:\mathrm{is}\:\mathrm{impossible}\:\mathrm{since}\:\mathrm{101}\:\mathrm{is}\:\mathrm{prime} \\ $$$$\mathrm{Therefore}\:,\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:\mathrm{are}\:\mathrm{coprime}.\mathrm{From} \\ $$$$\mathrm{p}^{\mathrm{2}} =\mathrm{mn}\:\mathrm{and}\:\left(\mathrm{m},\mathrm{n}\right)=\mathrm{1}\:\mathrm{we}\:\mathrm{infer}\: \\ $$$$\mathrm{m}=\mathrm{u}^{\mathrm{2}} ,\mathrm{n}=\mathrm{v}^{\mathrm{2}} \mathrm{where}\:\mathrm{u}\:\mathrm{and}\:\mathrm{v}−\mathrm{coprime}, \\ $$$$\mathrm{u}>\mathrm{v}\:>\mathrm{0}\:\mathrm{and}\:\mathrm{are}\:\mathrm{odd}\:\mathrm{numbers}.\mathrm{Finally}, \\ $$$$\mathrm{we}\:\mathrm{obtain}\:\begin{cases}{\mathrm{p}=\mathrm{uv}\left(\mathrm{1}\right)}\\{\mathrm{q}=\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\right)}\\{\mathrm{101}=\frac{\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{3}\right)}\end{cases} \\ $$$$\left(\mathrm{3}\right)\Leftrightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{202}\Rightarrow\mathrm{v}<\mathrm{11}\leqslant\mathrm{u}\leqslant\mathrm{13} \\ $$$$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{u}=\mathrm{13}\Rightarrow\mathrm{v}^{\mathrm{2}} =\mathrm{202}−\mathrm{169}=\mathrm{33}\Rightarrow\nexists\mathrm{v}\in\mathbb{Z} \\ $$$$\left.\mathrm{ii}\right)\mathrm{For}\:\mathrm{u}=\mathrm{11}\Rightarrow\mathrm{v}^{\mathrm{2}} =\mathrm{202}−\mathrm{121}=\mathrm{81} \\ $$$$\Rightarrow\mathrm{v}=\mathrm{9}.\mathrm{Put}\:\mathrm{into}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{p},\mathrm{q}\right)\in\left(\mathrm{99},\mathrm{20}\right)\mathrm{and}\:\mathrm{due}\:\mathrm{to}\:\mathrm{above}\:\mathrm{note}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{equality}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{in}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{p},\mathrm{q}\right)\in\left\{\mathrm{99},\mathrm{20}\right),\left(−\mathrm{99},\mathrm{20}\right),\left(\mathrm{20},\mathrm{99}\right),\left(\mathrm{20},−\mathrm{99}\right), \\ $$$$\left.\left(\mathrm{99},−\mathrm{20}\right),\left(−\mathrm{99},−\mathrm{20}\right),\left(−\mathrm{20},\mathrm{99}\right),\left(−\mathrm{20},−\mathrm{99}\right)\right\} \\ $$

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