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Question Number 46809 by MJS last updated on 31/Oct/18

$$\mathrm{an}\mathrm{early}\mathrm{question}\mathrm{for}\mathrm{the}\mathrm{new}\mathrm{year}$$$$\mathrm{how}\mathrm{many}\mathrm{rectangular}\mathrm{triangles}\mathrm{with}\mathrm{sides}$$$$a,b,c\in {\mathbb{N}}^{★}\mathrm{exist}\mathrm{with}\mathrm{one}\mathrm{side}=2019$$

Answered by MrW3 last updated on 02/Nov/18

$$a^2+b^2=c^2$$$$\mathit{c}\mathit{a}\mathit{s}\mathit{e}1:withc=2019$$$$a^2+b^2={2019}^2=3^2\times {673}^2$$$$leta=3p,b=3q$$$$\Rightarrow p^2+q^2={673}^2$$$$ithasoneandonlyonesolution:$$$$p=385,q=552$$$$\Rightarrow a=3\times 385=1155$$$$\Rightarrow b=3\times 552=1656$$$$$$$$\mathit{c}\mathit{a}\mathit{s}\mathit{e}2:witha=2019$$$${2019}^2+b^2=c^2$$$$\Rightarrow c^2-b^2=(c-b)(c+b)={2019}^2=3^2\times {673}^2=m\times n=1\times 4076361=3\times 1358787=9\times 452929=673\times 6057$$$$c-b=m$$$$c+b=n$$$$\Rightarrow b=\frac{n-m}{2}$$$$\Rightarrow c=\frac{n+m}{2}$$$$\Rightarrow b=\frac{4076361-1}{2}=2038180$$$$\Rightarrow c=\frac{4076361+1}{2}=2038181$$$$or$$$$\Rightarrow b=\frac{1358787-3}{2}=679392$$$$\Rightarrow c=\frac{1358787+3}{2}=679395$$$$or$$$$\Rightarrow b=\frac{452929-9}{2}=226460$$$$\Rightarrow c=\frac{452929+9}{2}=226469$$$$or$$$$\Rightarrow b=\frac{6057-673}{2}=2692$$$$\Rightarrow c=\frac{6057+673}{2}=3365$$$$$$$$so\mathit{t}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathit{l}\mathit{y}thereexist5suchtriangles:$$$$a/b/c=1155/1656/2019$$$$a/b/c=2019/2692/3365$$$$a/b/c=2019/226460/226469$$$$a/b/c=2019/679392/679395$$$$a/b/c=2019/2038180/2038181$$

Commented by MJS last updated on 02/Nov/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right},\mathrm{thank}\mathrm{you}!$$

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