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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

$${\mathrm{p}}^2+7\mathrm{pq}+{\mathrm{q}}^2={\mathrm{n}}^2;\mathrm{n}\in \mathbb{Z}$$$${\mathrm{p}}^2+7\mathrm{pq}+{\mathrm{q}}^2-{\mathrm{n}}^2=0$$$$\mathrm{p}=\frac{-7\mathrm{q}\pm \sqrt{{49\mathrm{q}}^2-4({\mathrm{q}}^2-{\mathrm{n}}^2)}}{2}$$$$=\frac{-7\mathrm{q}\pm \sqrt{{45\mathrm{q}}^2+{4\mathrm{n}}^2}}{2}$$$${45\mathrm{q}}^2+{4\mathrm{n}}^2\mathrm{is}\mathrm{necessarily}\mathrm{perfect}\mathrm{square}$$$$\mathrm{Let}{45\mathrm{q}}^2+{4\mathrm{n}}^2={\mathrm{m}}^2$$$${\mathrm{q}}^2=\frac{{\mathrm{m}}^2-{4\mathrm{n}}^2}{45}=\frac{(\mathrm{m}-2\mathrm{n})(\mathrm{m}+2\mathrm{n})}{3^2.5}$$$$\mathrm{possible}\mathrm{cases}\mathrm{of}\mathrm{writing}{\mathrm{q}}^2\mathrm{as}\mathrm{product}$$$$\mathrm{of}\mathrm{two}\mathrm{factors}:$$$${\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{3}\times \frac{\mathrm{m}+2\mathrm{n}}{15}.......\left(\mathrm{i}\right)$$$${\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{9}\times \frac{\mathrm{m}+2\mathrm{n}}{5}.........\left(\mathrm{ii}\right)$$$${\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{15}\times \frac{\mathrm{m}+2\mathrm{n}}{3}........\left(\mathrm{iii}\right)$$$${\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{45}\times \frac{\mathrm{m}+2\mathrm{n}}{1}.......\left(\mathrm{iv}\right)$$$${\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{1}\times \frac{\mathrm{m}+2\mathrm{n}}{45}........\left(\mathrm{v}\right)$$$$\left(\mathrm{i}\right){\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{3}\times \frac{\mathrm{m}+2\mathrm{n}}{15}$$$$\mathrm{As}\mathrm{q}\mathrm{is}\mathrm{prime}\mathrm{so}\mathrm{there}\mathrm{are}3\mathrm{possible}$$$$\mathrm{cases}:$$$$\mathrm{Case}-1\mathrm{q}=\frac{\mathrm{m}-2\mathrm{n}}{3}=\frac{\mathrm{m}+2\mathrm{n}}{15}$$$$15\mathrm{m}-30\mathrm{n}=3\mathrm{m}+6\mathrm{n}$$$$\mathrm{m}=3\mathrm{n}$$$$\mathrm{q}=\frac{\mathrm{n}}{3}\Rightarrow \mathrm{n}=3\Rightarrow \mathrm{q}=1\notin \mathbb{P}$$$$\mathrm{n}=6\Rightarrow \mathrm{q}=2$$$$\mathrm{p}=\frac{-7\mathrm{q}\pm \sqrt{{45\mathrm{q}}^2+{4\mathrm{n}}^2}}{2}$$$$$$$$\mathrm{p}=\frac{-7\left(2\right)\pm \sqrt{45{\left(2\right)}^2+4{\left(6\right)}^2}}{2}$$$$\mathrm{p}=\frac{-21\pm \sqrt{180+144}}{2}=\frac{-21\pm 18}{2}\notin \mathbb{P}$$$$\mathrm{Case}-2\frac{\mathrm{m}-2\mathrm{n}}{3}=1\wedge {\mathrm{q}}^2=\frac{\mathrm{m}+2\mathrm{n}}{15}$$$$\mathrm{m}=3+2\mathrm{n}\Rightarrow {\mathrm{q}}^2=\frac{3+4\mathrm{n}}{15}$$$$4\mathrm{n}+3\equiv 0\left(\mathrm{mod}15\right)$$$$4\mathrm{n}\equiv 12\left(\mathrm{mod}15\right)$$$$\mathrm{n}\equiv 3\left(\mathrm{mod}15\right)$$$$\mathrm{n}=3,18,33,48,...93$$$$\mathrm{q}=3$$$$\mathrm{q}=5$$$$\mathrm{Case}-3{\mathrm{q}}^2=\frac{\mathrm{m}-2\mathrm{n}}{3}\wedge \frac{\mathrm{m}+2\mathrm{n}}{15}=1$$$$⋮$$

Answered by Rasheed.Sindhi last updated on 17/Jan/18

$$\mathrm{Let}{\mathrm{p}}^2+7\mathrm{pq}+{\mathrm{q}}^2={\mathrm{n}}^2;\mathrm{n}\in \mathbb{Z}$$$${\mathrm{p}}^2+7\mathrm{pq}+{\mathrm{q}}^2-{\mathrm{n}}^2=0$$$$\mathrm{p}=\frac{-7\mathrm{q}\pm \sqrt{{49\mathrm{q}}^2-4({\mathrm{q}}^2-{\mathrm{n}}^2)}}{2}$$$$\mathrm{p}=\frac{-7\mathrm{q}\pm \sqrt{{45\mathrm{q}}^2+{4\mathrm{n}}^2}}{2}$$$${45\mathrm{q}}^2+{4\mathrm{n}}^2\mathrm{is}\mathrm{necessarily}\mathrm{perfect}\mathrm{square}$$$$\mathrm{Let}{45\mathrm{q}}^2+{4\mathrm{n}}^2={\mathrm{m}}^2;\mathrm{m}\in \mathbb{Z}$$$${45\mathrm{q}}^2={\mathrm{m}}^2-{4\mathrm{n}}^2$$$${45\mathrm{q}}^2=(\mathrm{m}-2\mathrm{n})(\mathrm{m}+2\mathrm{n})$$$$3^2.5.{\mathrm{q}}^2=(\mathrm{m}-2\mathrm{n})(\mathrm{m}+2\mathrm{n})$$$$\mathrm{Possible}\mathrm{cases}:$$$$\bullet \mathrm{m}-2\mathrm{n}=3\Rightarrow \mathrm{m}+2\mathrm{n}={15\mathrm{q}}^2$$$$\mathrm{m}=2\mathrm{n}+3\Rightarrow (2\mathrm{n}+3)+2\mathrm{n}={15\mathrm{q}}^2$$$${\mathrm{q}}^2=\frac{4\mathrm{n}+3}{15}\Rightarrow 15\mid (4\mathrm{n}+3)$$$$4\mathrm{n}+3\equiv 0\left(\mathrm{mod}15\right)$$$$4\mathrm{n}\equiv -3+15\left(\mathrm{mod}15\right)$$$$\mathrm{n}\equiv 3\left(\mathrm{mod}15\right)$$$$\mathrm{n}=3,18,33,48,....$$$$\mathrm{But}\mathrm{we}\mathrm{need}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}\mathrm{for}$$$$\mathrm{which}\frac{4\mathrm{n}+3}{15}\mathrm{is}\mathrm{perfect}\mathrm{square}\mathrm{of}$$$$\mathrm{any}\mathrm{prime}.\mathrm{Testing}\mathrm{the}\mathrm{values}\mathrm{one}$$$$\mathrm{by}\mathrm{one}\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{n}=93\mathrm{fulfill}\mathrm{our}$$$$\mathrm{criteria}.\frac{4\left(93\right)+3}{15}=\frac{375}{15}=25$$$$\therefore \mathrm{q}=5$$$$\mathrm{But}\mathrm{for}\mathrm{q}=5$$$$\mathrm{p}=\frac{-7\left(5\right)\pm \sqrt{45{\left(5\right)}^2+4{\left(93\right)}^2}}{2}=77\notin \mathbb{P}$$$$\bullet \mathrm{m}-2\mathrm{n}=5\Rightarrow \mathrm{m}+2\mathrm{n}={9\mathrm{q}}^2$$$$\bullet \mathrm{m}-2\mathrm{n}=9\Rightarrow \mathrm{m}+2\mathrm{n}={5\mathrm{q}}^2$$$$\bullet \mathrm{m}-2\mathrm{n}=15\Rightarrow \mathrm{m}+2\mathrm{n}={3\mathrm{q}}^2$$$$\bullet \mathrm{m}-2\mathrm{n}=3\mathrm{q}\Rightarrow \mathrm{m}+2\mathrm{n}=15\mathrm{q}$$$$\bullet \mathrm{m}-2\mathrm{n}=5\mathrm{q}\Rightarrow \mathrm{m}+2\mathrm{n}=9\mathrm{q}$$$$\bullet \mathrm{m}-2\mathrm{n}=9\mathrm{q}\Rightarrow \mathrm{m}+2\mathrm{n}=5\mathrm{q}$$$$\bullet \mathrm{m}-2\mathrm{n}=15\mathrm{q}\Rightarrow \mathrm{m}+2\mathrm{n}=3\mathrm{q}$$$$\bullet \mathrm{m}-2\mathrm{n}={3\mathrm{q}}^2\Rightarrow \mathrm{m}+2\mathrm{n}=15$$$$\bullet \mathrm{m}-2\mathrm{n}={5\mathrm{q}}^2\Rightarrow \mathrm{m}+2\mathrm{n}=9$$$$\bullet \mathrm{m}-2\mathrm{n}={9\mathrm{q}}^2\Rightarrow \mathrm{m}+2\mathrm{n}=5$$$$\bullet \mathrm{m}-2\mathrm{n}={15\mathrm{q}}^2\Rightarrow \mathrm{m}+2\mathrm{n}=3$$$$\mathrm{Continue}$$$$$$

Answered by mrW2 last updated on 18/Jan/18

$$Ifpandqmaybeequal,theneveryprime$$$$isasolution,i.e.p=q=1,2,3,5,7,...$$$$$$$$Ifpandqshouldbedistinct,thenthereis$$$$onlyonesolution:$$$$p=3,q=11orp=11,q=3.$$

Commented by mrW2 last updated on 18/Jan/18

$$Intuitivelywecansee,ifq=pweget$$$$p^2+7pq+q^2=9p^2={\left(3p\right)}^2=n^2$$$$thisisalwaystrueforeveryprimep.$$$$Butwecanalsogetthisresultbelow$$$$withoutusingintuition.$$$$$$$$p^2+7pq+q^2=n^2$$$$\Rightarrow p^2+2pq+q^2+5pq=n^2$$$$\Rightarrow n^2-{(p+q)}^2=5pq$$$$\Rightarrow (n-p-q)(n+p+q)=5pq$$$$$$$$sincepandqareprimes,thereare$$$$onlyfollowingcases:$$$$case1:$$$$\{\begin{array}{l}n-p-q=5p\\ n+p+q=q\end{array}$$$$\Rightarrow 5p+2p+2q=q$$$$\Rightarrow 7p=-q<0\Rightarrow impossible$$$$case2:$$$$\{\begin{array}{l}n-p-q=5q\\ n+p+q=p\end{array}$$$$\Rightarrow asaboveimpossible$$$$$$$$case3:$$$$\{\begin{array}{l}n-p-q=p\\ n+p+q=5q\end{array}$$$$\Rightarrow p+2p+2q=5q$$$$\Rightarrow p=q=everyprime$$$$case4:$$$$\{\begin{array}{l}n-p-q=q\\ n+p+q=5p\end{array}$$$$\Rightarrow q+2p+2q=5p$$$$\Rightarrow q=p=everyprime$$$$$$$$case5:$$$$\{\begin{array}{l}n-p-q=5\\ n+p+q=pq\end{array}$$$$\Rightarrow 5+2p+2q=pq$$$$\Rightarrow q=\frac{2p+5}{p-2}=2+\frac{9}{p-2}$$$$\Rightarrow p=3,q=11$$$$\Rightarrow p=5,q=5$$$$\Rightarrow p=11,q=3$$$$case6:$$$$\{\begin{array}{l}n-p-q=pq\\ n+p+q=5\end{array}$$$$\Rightarrow pq+2p+2q=5$$$$\Rightarrow q=\frac{5-2p}{p+2}=\frac{9}{p+2}-2$$$$\Rightarrow p=1,q=1$$$$\Rightarrow p=7,q=-1<0\Rightarrow impossible$$$$$$$$Thereforethesolutionsare:$$$$p=q=everyprime$$$$p=3,q=11$$$$p=11,q=3$$

Commented by mrW2 last updated on 18/Jan/18

$$Thankyousir!$$$$Buthowdidyouwritethis?$$

Commented by Tinkutara last updated on 18/Jan/18

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Commented by Rasheed.Sindhi last updated on 18/Jan/18

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Commented by Rasheed.Sindhi last updated on 18/Jan/18

$$Sirthisisa\mathit{p}\mathit{l}\mathit{a}\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{x}\mathit{t}\mathit{c}\mathit{o}\mathit{m}\mathit{m}\mathit{e}\mathit{n}\mathit{t}.$$$$Ihavepasteditfrom{}^{\prime }\mathrm{stylish}\mathrm{text}{}^{\prime }$$$$app.$$

Commented by mrW2 last updated on 18/Jan/18

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