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Question Number 27936 by Tinkutara last updated on 17/Jan/18
Commented by Rasheed.Sindhi last updated on 19/Jan/18
p^2 +7pq+q^2 =n^2 ; n∈Z p^2 +7pq+q^2 −n^2 =0 p=((−7q±(√(49q^2 −4(q^2 −n^2 ))))/2) =((−7q±(√(45q^2 +4n^2 )))/2) 45q^2 +4n^2 is necessarily perfect square Let 45q^2 +4n^2 =m^2 q^2 =((m^2 −4n^2 )/(45))=(((m−2n)(m+2n))/(3^2 .5)) possible cases of writing q^2 as product of two factors: q^2 =((m−2n)/3)×((m+2n)/(15)) .......(i) q^2 =((m−2n)/9)×((m+2n)/5).........(ii) q^2 =((m−2n)/(15))×((m+2n)/3) ........(iii) q^2 =((m−2n)/(45))×((m+2n)/1) .......(iv) q^2 =((m−2n)/1)×((m+2n)/(45))........(v) (i)q^2 = ((m−2n)/3)×((m+2n)/(15)) As q is prime so there are 3 possible cases: Case-1 q=((m−2n)/3)=((m+2n)/(15)) 15m−30n=3m+6n m=3n q=(n/3)⇒n=3 ⇒q=1∉P n=6⇒q=2 p=((−7q±(√(45q^2 +4n^2 )))/2) p=((−7(2)±(√(45(2)^2 +4(6)^2 )))/2) p=((−21±(√(180+144)))/2)=((−21±18)/2)∉P Case-2 ((m−2n)/3)=1∧ q^2 =((m+2n)/(15)) m=3+2n⇒q^2 =((3+4n)/(15)) 4n+3≡0(mod 15) 4n≡12(mod 15) n≡3(mod 15) n=3,18,33,48,...93 q=3 q=5 Case-3 q^2 =((m−2n)/3)∧ ((m+2n)/(15))=1 ⋮
$$\mathrm{p}^{\mathrm{2}} +\mathrm{7pq}+\mathrm{q}^{\mathrm{2}} =\mathrm{n}^{\mathrm{2}} \:\:;\:\mathrm{n}\in\mathbb{Z} \\ $$$$\mathrm{p}^{\mathrm{2}} +\mathrm{7pq}+\mathrm{q}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{p}=\frac{−\mathrm{7q}\pm\sqrt{\mathrm{49q}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{q}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\:\:\:=\frac{−\mathrm{7q}\pm\sqrt{\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} \:\mathrm{is}\:\:\mathrm{necessarily}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\:\mathrm{Let}\:\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} \\ $$$$\:\:\mathrm{q}^{\mathrm{2}} =\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{4n}^{\mathrm{2}} }{\mathrm{45}}=\frac{\left(\mathrm{m}−\mathrm{2n}\right)\left(\mathrm{m}+\mathrm{2n}\right)}{\mathrm{3}^{\mathrm{2}} .\mathrm{5}} \\ $$$$\mathrm{possible}\:\mathrm{cases}\:\mathrm{of}\:\mathrm{writing}\:\mathrm{q}^{\mathrm{2}} \mathrm{as}\:\mathrm{product} \\ $$$$\mathrm{of}\:\mathrm{two}\:\mathrm{factors}:\: \\ $$$$\:\:\:\:\:\:\mathrm{q}^{\mathrm{2}} \:=\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{3}}×\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{15}}\:…….\left(\mathrm{i}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{q}^{\mathrm{2}} \:=\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{9}}×\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{5}}………\left(\mathrm{ii}\right)\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{q}^{\mathrm{2}} \:=\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{15}}×\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{3}}\:……..\left(\mathrm{iii}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{q}^{\mathrm{2}} \:=\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{45}}×\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{1}}\:…….\left(\mathrm{iv}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{q}^{\mathrm{2}} \:=\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{1}}×\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{45}}……..\left(\mathrm{v}\right)\:\:\:\: \\ $$$$\left(\mathrm{i}\right)\mathrm{q}^{\mathrm{2}} =\:\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{3}}×\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{15}} \\ $$$$\mathrm{As}\:\mathrm{q}\:\mathrm{is}\:\mathrm{prime}\:\mathrm{so}\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{possible} \\ $$$$\mathrm{cases}: \\ $$$$\:\mathrm{Case}-\mathrm{1}\:\:\mathrm{q}=\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{3}}=\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{15}}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{15m}−\mathrm{30n}=\mathrm{3m}+\mathrm{6n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{m}=\mathrm{3n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{q}=\frac{\mathrm{n}}{\mathrm{3}}\Rightarrow\mathrm{n}=\mathrm{3}\:\Rightarrow\mathrm{q}=\mathrm{1}\notin\mathbb{P} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}=\mathrm{6}\Rightarrow\mathrm{q}=\mathrm{2} \\ $$$$\:\:\mathrm{p}=\frac{−\mathrm{7q}\pm\sqrt{\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\mathrm{p}=\frac{−\mathrm{7}\left(\mathrm{2}\right)\pm\sqrt{\mathrm{45}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{6}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:\mathrm{p}=\frac{−\mathrm{21}\pm\sqrt{\mathrm{180}+\mathrm{144}}}{\mathrm{2}}=\frac{−\mathrm{21}\pm\mathrm{18}}{\mathrm{2}}\notin\mathbb{P} \\ $$$$\:\mathrm{Case}-\mathrm{2}\:\:\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{3}}=\mathrm{1}\wedge\:\mathrm{q}^{\mathrm{2}} =\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{15}}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{m}=\mathrm{3}+\mathrm{2n}\Rightarrow\mathrm{q}^{\mathrm{2}} =\frac{\mathrm{3}+\mathrm{4n}}{\mathrm{15}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4n}+\mathrm{3}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4n}\equiv\mathrm{12}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\:\mathrm{n}=\mathrm{3},\mathrm{18},\mathrm{33},\mathrm{48},…\mathrm{93} \\ $$$$\:\:\:\:\mathrm{q}=\mathrm{3} \\ $$$$\:\:\:\:\:\mathrm{q}=\mathrm{5} \\ $$$$\:\mathrm{Case}-\mathrm{3}\:\:\mathrm{q}^{\mathrm{2}} =\frac{\mathrm{m}−\mathrm{2n}}{\mathrm{3}}\wedge\:\frac{\mathrm{m}+\mathrm{2n}}{\mathrm{15}}=\mathrm{1}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\vdots \\ $$
Answered by Rasheed.Sindhi last updated on 17/Jan/18
Let p^2 +7pq+q^2 =n^2 ; n∈Z p^2 +7pq+q^2 −n^2 =0 p=((−7q±(√(49q^2 −4(q^2 −n^2 ))))/2) p=((−7q±(√(45q^2 +4n^2 )))/2) 45q^2 +4n^2 is necessarily perfect square Let 45q^2 +4n^2 =m^2 ;m∈Z 45q^2 =m^2 −4n^2 45q^2 =(m−2n)(m+2n) 3^2 .5.q^2 =(m−2n)(m+2n) Possible cases: •m−2n=3⇒m+2n=15q^2 m=2n+3⇒(2n+3)+2n=15q^2 q^2 =((4n+3)/(15)) ⇒15∣(4n+3) 4n+3≡0(mod 15) 4n≡−3+15(mod 15) n≡3(mod15) n=3,18,33,48,.... But we need the value of n for which ((4n+3)/(15)) is perfect square of any prime.Testing the values one by one we see that n=93 fulfill our criteria. ((4(93)+3)/(15))=((375)/(15))=25 ∴q=5 But for q=5 p=((−7(5)±(√(45(5)^2 +4(93)^2 )))/2)=77∉P •m−2n=5⇒m+2n=9q^2 •m−2n=9⇒m+2n=5q^2 •m−2n=15⇒m+2n=3q^2 •m−2n=3q⇒m+2n=15q •m−2n=5q⇒m+2n=9q •m−2n=9q⇒m+2n=5q •m−2n=15q⇒m+2n=3q •m−2n=3q^2 ⇒m+2n=15 •m−2n=5q^2 ⇒m+2n=9 •m−2n=9q^2 ⇒m+2n=5 •m−2n=15q^2 ⇒m+2n=3 Continue
$$\mathrm{Let}\:\mathrm{p}^{\mathrm{2}} +\mathrm{7pq}+\mathrm{q}^{\mathrm{2}} =\mathrm{n}^{\mathrm{2}} \:;\:\:\mathrm{n}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:\mathrm{p}^{\mathrm{2}} +\mathrm{7pq}+\mathrm{q}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{p}=\frac{−\mathrm{7q}\pm\sqrt{\mathrm{49q}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{q}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{p}=\frac{−\mathrm{7q}\pm\sqrt{\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{necessarily}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\mathrm{Let}\:\mathrm{45q}^{\mathrm{2}} +\mathrm{4n}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} \:\:;\mathrm{m}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{45q}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} −\mathrm{4n}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{45q}^{\mathrm{2}} =\left(\mathrm{m}−\mathrm{2n}\right)\left(\mathrm{m}+\mathrm{2n}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{2}} .\mathrm{5}.\mathrm{q}^{\mathrm{2}} =\left(\mathrm{m}−\mathrm{2n}\right)\left(\mathrm{m}+\mathrm{2n}\right) \\ $$$$\mathrm{Possible}\:\mathrm{cases}: \\ $$$$\:\bullet\mathrm{m}−\mathrm{2n}=\mathrm{3}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{15q}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{m}=\mathrm{2n}+\mathrm{3}\Rightarrow\left(\mathrm{2n}+\mathrm{3}\right)+\mathrm{2n}=\mathrm{15q}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{q}^{\mathrm{2}} =\frac{\mathrm{4n}+\mathrm{3}}{\mathrm{15}}\:\Rightarrow\mathrm{15}\mid\left(\mathrm{4n}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4n}+\mathrm{3}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4n}\equiv−\mathrm{3}+\mathrm{15}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{n}\equiv\mathrm{3}\left(\mathrm{mod15}\right) \\ $$$$\:\:\:\mathrm{n}=\mathrm{3},\mathrm{18},\mathrm{33},\mathrm{48},…. \\ $$$$\mathrm{But}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n}\:\mathrm{for} \\ $$$$\mathrm{which}\:\frac{\mathrm{4n}+\mathrm{3}}{\mathrm{15}}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{of}\: \\ $$$$\mathrm{any}\:\mathrm{prime}.\mathrm{Testing}\:\mathrm{the}\:\mathrm{values}\:\mathrm{one} \\ $$$$\mathrm{by}\:\mathrm{one}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{n}=\mathrm{93}\:\mathrm{fulfill}\:\mathrm{our} \\ $$$$\mathrm{criteria}.\:\frac{\mathrm{4}\left(\mathrm{93}\right)+\mathrm{3}}{\mathrm{15}}=\frac{\mathrm{375}}{\mathrm{15}}=\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\mathrm{q}=\mathrm{5} \\ $$$$\mathrm{But}\:\mathrm{for}\:\mathrm{q}=\mathrm{5}\: \\ $$$$\:\:\:\mathrm{p}=\frac{−\mathrm{7}\left(\mathrm{5}\right)\pm\sqrt{\mathrm{45}\left(\mathrm{5}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{93}\right)^{\mathrm{2}} }}{\mathrm{2}}=\mathrm{77}\notin\mathbb{P} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{5}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{9q}^{\mathrm{2}} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{9}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{5q}^{\mathrm{2}} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{15}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{3q}^{\mathrm{2}} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{3q}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{15q} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{5q}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{9q} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{9q}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{5q} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{15q}\Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{3q} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{3q}^{\mathrm{2}} \Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{15} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{5q}^{\mathrm{2}} \Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{9} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{9q}^{\mathrm{2}} \Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{5} \\ $$$$\bullet\mathrm{m}−\mathrm{2n}=\mathrm{15q}^{\mathrm{2}} \Rightarrow\mathrm{m}+\mathrm{2n}=\mathrm{3} \\ $$$$\mathrm{Continue} \\ $$$$ \\ $$
Answered by mrW2 last updated on 18/Jan/18
If p and q may be equal, then every prime is a solution, i.e. p=q=1,2,3,5,7,... If p and q should be distinct, then there is only one solution: p=3, q=11 or p=11, q=3.
$${If}\:{p}\:{and}\:{q}\:{may}\:{be}\:{equal},\:{then}\:{every}\:{prime} \\ $$$${is}\:{a}\:{solution},\:{i}.{e}.\:{p}={q}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7},… \\ $$$$ \\ $$$${If}\:{p}\:{and}\:{q}\:{should}\:{be}\:{distinct},\:{then}\:{there}\:{is} \\ $$$${only}\:{one}\:{solution}: \\ $$$${p}=\mathrm{3},\:{q}=\mathrm{11}\:{or}\:{p}=\mathrm{11},\:{q}=\mathrm{3}. \\ $$
Commented by mrW2 last updated on 18/Jan/18
Intuitively we can see, if q=p we get p^2 +7pq+q^2 =9p^2 =(3p)^2 =n^2 this is always true for every prime p. But we can also get this result below without using intuition. p^2 +7pq+q^2 =n^2 ⇒p^2 +2pq+q^2 +5pq=n^2 ⇒n^2 −(p+q)^2 =5pq ⇒(n−p−q)(n+p+q)=5pq since p and q are primes, there are only following cases: case 1: { ((n−p−q=5p)),((n+p+q=q)) :} ⇒5p+2p+2q=q ⇒7p=−q<0 ⇒impossible case 2: { ((n−p−q=5q)),((n+p+q=p)) :} ⇒as above impossible case 3: { ((n−p−q=p)),((n+p+q=5q)) :} ⇒p+2p+2q=5q ⇒p=q=every prime case 4: { ((n−p−q=q)),((n+p+q=5p)) :} ⇒q+2p+2q=5p ⇒q=p=every prime case 5: { ((n−p−q=5)),((n+p+q=pq)) :} ⇒5+2p+2q=pq ⇒q=((2p+5)/(p−2))=2+(9/(p−2)) ⇒p=3, q=11 ⇒p=5, q=5 ⇒p=11, q=3 case 6: { ((n−p−q=pq)),((n+p+q=5)) :} ⇒pq+2p+2q=5 ⇒q=((5−2p)/(p+2))=(9/(p+2))−2 ⇒p=1, q=1 ⇒p=7, q=−1<0 ⇒impossible Therefore the solutions are: p=q=every prime p=3, q=11 p=11, q=3
$${Intuitively}\:{we}\:{can}\:{see},\:{if}\:{q}={p}\:{we}\:{get} \\ $$$${p}^{\mathrm{2}} +\mathrm{7}{pq}+{q}^{\mathrm{2}} =\mathrm{9}{p}^{\mathrm{2}} =\left(\mathrm{3}{p}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$$${this}\:{is}\:{always}\:{true}\:{for}\:{every}\:{prime}\:{p}. \\ $$$${But}\:{we}\:{can}\:{also}\:{get}\:{this}\:{result}\:{below} \\ $$$${without}\:{using}\:{intuition}. \\ $$$$ \\ $$$${p}^{\mathrm{2}} +\mathrm{7}{pq}+{q}^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +\mathrm{2}{pq}+{q}^{\mathrm{2}} +\mathrm{5}{pq}={n}^{\mathrm{2}} \\ $$$$\Rightarrow{n}^{\mathrm{2}} −\left({p}+{q}\right)^{\mathrm{2}} =\mathrm{5}{pq} \\ $$$$\Rightarrow\left({n}−{p}−{q}\right)\left({n}+{p}+{q}\right)=\mathrm{5}{pq} \\ $$$$ \\ $$$${since}\:{p}\:{and}\:{q}\:{are}\:{primes},\:{there}\:{are} \\ $$$${only}\:{following}\:{cases}: \\ $$$${case}\:\mathrm{1}: \\ $$$$\begin{cases}{{n}−{p}−{q}=\mathrm{5}{p}}\\{{n}+{p}+{q}={q}}\end{cases} \\ $$$$\Rightarrow\mathrm{5}{p}+\mathrm{2}{p}+\mathrm{2}{q}={q} \\ $$$$\Rightarrow\mathrm{7}{p}=−{q}<\mathrm{0}\:\Rightarrow{impossible} \\ $$$${case}\:\mathrm{2}: \\ $$$$\begin{cases}{{n}−{p}−{q}=\mathrm{5}{q}}\\{{n}+{p}+{q}={p}}\end{cases} \\ $$$$\Rightarrow{as}\:{above}\:{impossible} \\ $$$$ \\ $$$${case}\:\mathrm{3}: \\ $$$$\begin{cases}{{n}−{p}−{q}={p}}\\{{n}+{p}+{q}=\mathrm{5}{q}}\end{cases} \\ $$$$\Rightarrow{p}+\mathrm{2}{p}+\mathrm{2}{q}=\mathrm{5}{q} \\ $$$$\Rightarrow{p}={q}={every}\:{prime} \\ $$$${case}\:\mathrm{4}: \\ $$$$\begin{cases}{{n}−{p}−{q}={q}}\\{{n}+{p}+{q}=\mathrm{5}{p}}\end{cases} \\ $$$$\Rightarrow{q}+\mathrm{2}{p}+\mathrm{2}{q}=\mathrm{5}{p} \\ $$$$\Rightarrow{q}={p}={every}\:{prime} \\ $$$$ \\ $$$${case}\:\mathrm{5}: \\ $$$$\begin{cases}{{n}−{p}−{q}=\mathrm{5}}\\{{n}+{p}+{q}={pq}}\end{cases} \\ $$$$\Rightarrow\mathrm{5}+\mathrm{2}{p}+\mathrm{2}{q}={pq} \\ $$$$\Rightarrow{q}=\frac{\mathrm{2}{p}+\mathrm{5}}{{p}−\mathrm{2}}=\mathrm{2}+\frac{\mathrm{9}}{{p}−\mathrm{2}} \\ $$$$\Rightarrow{p}=\mathrm{3},\:{q}=\mathrm{11} \\ $$$$\Rightarrow{p}=\mathrm{5},\:{q}=\mathrm{5} \\ $$$$\Rightarrow{p}=\mathrm{11},\:{q}=\mathrm{3} \\ $$$${case}\:\mathrm{6}: \\ $$$$\begin{cases}{{n}−{p}−{q}={pq}}\\{{n}+{p}+{q}=\mathrm{5}}\end{cases} \\ $$$$\Rightarrow{pq}+\mathrm{2}{p}+\mathrm{2}{q}=\mathrm{5} \\ $$$$\Rightarrow{q}=\frac{\mathrm{5}−\mathrm{2}{p}}{{p}+\mathrm{2}}=\frac{\mathrm{9}}{{p}+\mathrm{2}}−\mathrm{2} \\ $$$$\Rightarrow{p}=\mathrm{1},\:{q}=\mathrm{1} \\ $$$$\Rightarrow{p}=\mathrm{7},\:{q}=−\mathrm{1}<\mathrm{0}\:\Rightarrow{impossible} \\ $$$$ \\ $$$${Therefore}\:{the}\:{solutions}\:{are}: \\ $$$${p}={q}={every}\:{prime} \\ $$$${p}=\mathrm{3},\:{q}=\mathrm{11} \\ $$$${p}=\mathrm{11},\:{q}=\mathrm{3} \\ $$
Commented by mrW2 last updated on 18/Jan/18
Thank you sir! But how did you write this?
$${Thank}\:{you}\:{sir}! \\ $$$${But}\:{how}\:{did}\:{you}\:{write}\:{this}? \\ $$
Commented by Tinkutara last updated on 18/Jan/18
ᝨℍᗅℕK ℽᝪႮ Ꮙℰℛℽ ℳႮℂℍ Տⅈℛ��! ⅈ ℊᝪᝨ ᝨℍℰ ᗅℕՏᗯℰℛ.
Commented by Rasheed.Sindhi last updated on 18/Jan/18
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Commented by Rasheed.Sindhi last updated on 18/Jan/18
Sir this is a plaintext comment. I have pasted it from ′ stylish text ′ app.
$${Sir}\:{this}\:{is}\:{a}\:\boldsymbol{{plaintext}}\:\boldsymbol{{comment}}. \\ $$$${I}\:{have}\:{pasted}\:{it}\:{from}\:'\:\mathrm{stylish}\:\mathrm{text}\:'\: \\ $$$${app}. \\ $$
Commented by mrW2 last updated on 18/Jan/18
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