Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 165098 by mnjuly1970 last updated on 26/Jan/22

Answered by Eulerian last updated on 26/Jan/22

Solution: We know that: 1^2 +2^2 +3^2 +......+n^2 = ((n(n+1)(2n+1))/6) ∴ ((n(n+1)(2n+1))/6) = m^2 6m^2 = (n^2 +n)(2n+1) 6m^2 = 2n^3 +3n^2 +n 3m^2 = n^3 +((3n^2 )/2)+(n/2) One can easily notice that when (m, n) = (1, 1) the equation satisfies the condition. By transforming the right hand side into a depressed cubic polynomial, let: n = z − ((((3/2)))/3) = z − (1/2) ∴ (z − (1/2))^3 + (3/2)∙(z − (1/2))^2 + (1/2)∙(z − (1/2)) = 3m^2 z^3 + ((1/2) − ((((3/2))^2 )/3))∙z + ((2((3/2))^3 −9((3/2))((1/2)))/(27)) = 3m^2 z^3 − (z/4) = 3m^2 let: 3m^2 = y^2 Notice that we now have a form of elliptic curve equation: y^2 = z^3 −(z/4) And from our previous solution, it works at (y, z) = (±(√3) , (3/2)) . By building a tangent line to the elliptic curve, we have: y′ = (d/dz) ((√(z^3 −(z/4)))) = (((12z^2 −1)/(8z^3 −2z)))∙(√(z^3 −(z/4))) m_(tan) = ((13(√3))/(12)) ∴ y−y_1 = m_(tan) (z−z_1 ) y = ((13(√3))/(12))∙(z − (3/2)) + (√3) Thus, we can now see all of the possible solutions to the equation: ((13(√3))/(12))∙(z − (3/2)) + (√3) = (√(z^3 −(z/4))) z = ((25)/(48)) Therefore, we now have: y^2 = (((25)/(48)))^3 − ((1/4))(((25)/(48))) y = ± ((35(√3))/(576)) ∴ (y, z) = (± ((35(√3))/(576)), ((25)/(48))) , (±(√3) , (3/2)) By substituting back, we now have: 3m^2 = (± ((35(√3))/(576)))^2 m = ± ((35)/(576)) and ((25)/(48)) − (1/2) = n n = (1/(48)) ∴ (m, n) = (±1, 1), (± ((35)/(576)) , (1/(48))) By graphing the elliptic curve and its tangent line:

$$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{We}\:\mathrm{know}\:\mathrm{that}: \\ $$$$\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +......+\mathrm{n}^{\mathrm{2}} \:=\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\therefore \\ $$$$\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}\:=\:\mathrm{m}^{\mathrm{2}} \\ $$$$\:\mathrm{6m}^{\mathrm{2}} \:=\:\left(\mathrm{n}^{\mathrm{2}} +\mathrm{n}\right)\left(\mathrm{2n}+\mathrm{1}\right) \\ $$$$\:\mathrm{6m}^{\mathrm{2}} \:=\:\mathrm{2n}^{\mathrm{3}} +\mathrm{3n}^{\mathrm{2}} +\mathrm{n} \\ $$$$\:\mathrm{3m}^{\mathrm{2}} \:=\:\mathrm{n}^{\mathrm{3}} +\frac{\mathrm{3n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{n}}{\mathrm{2}} \\ $$$$\: \\ $$$$\:\mathrm{One}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{notice}\:\mathrm{that}\:\mathrm{when}\:\left(\mathrm{m},\:\mathrm{n}\right)\:=\:\left(\mathrm{1},\:\mathrm{1}\right) \\ $$$$\:\mathrm{the}\:\mathrm{equation}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{condition}.\:\mathrm{By}\:\mathrm{transforming} \\ $$$$\:\mathrm{the}\:\mathrm{right}\:\mathrm{hand}\:\mathrm{side}\:\mathrm{into}\:\mathrm{a}\:\mathrm{depressed}\:\mathrm{cubic}\:\mathrm{polynomial},\:\mathrm{let}: \\ $$$$\:\mathrm{n}\:=\:\mathrm{z}\:−\:\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{3}}\:=\:\mathrm{z}\:−\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\: \\ $$$$\:\therefore \\ $$$$\:\left(\mathrm{z}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \:+\:\frac{\mathrm{3}}{\mathrm{2}}\centerdot\left(\mathrm{z}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\centerdot\left(\mathrm{z}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{3m}^{\mathrm{2}} \\ $$$$\:\mathrm{z}^{\mathrm{3}} \:+\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{3}}\right)\centerdot\mathrm{z}\:+\:\frac{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{9}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{27}}\:=\:\mathrm{3m}^{\mathrm{2}} \\ $$$$\:\mathrm{z}^{\mathrm{3}} \:−\:\frac{\mathrm{z}}{\mathrm{4}}\:=\:\mathrm{3m}^{\mathrm{2}} \\ $$$$\: \\ $$$$\:\mathrm{let}: \\ $$$$\:\mathrm{3m}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{2}} \\ $$$$\: \\ $$$$\:\mathrm{Notice}\:\mathrm{that}\:\mathrm{we}\:\mathrm{now}\:\mathrm{have}\:\mathrm{a}\:\mathrm{form}\:\mathrm{of}\:\mathrm{elliptic}\:\mathrm{curve}\:\mathrm{equation}: \\ $$$$\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{z}}{\mathrm{4}} \\ $$$$\: \\ $$$$\:\mathrm{And}\:\mathrm{from}\:\mathrm{our}\:\mathrm{previous}\:\mathrm{solution},\:\mathrm{it}\:\mathrm{works}\:\mathrm{at}\:\left(\mathrm{y},\:\mathrm{z}\right)\:=\:\left(\pm\sqrt{\mathrm{3}}\:,\:\frac{\mathrm{3}}{\mathrm{2}}\right)\:. \\ $$$$\:\mathrm{By}\:\mathrm{building}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{to}\:\mathrm{the}\:\mathrm{elliptic}\:\mathrm{curve},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\: \\ $$$$\:\mathrm{y}'\:=\:\frac{\mathrm{d}}{\mathrm{dz}}\:\left(\sqrt{\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{z}}{\mathrm{4}}}\right)\:=\:\left(\frac{\mathrm{12z}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8z}^{\mathrm{3}} −\mathrm{2z}}\right)\centerdot\sqrt{\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{z}}{\mathrm{4}}} \\ $$$$\: \\ $$$$\:\mathrm{m}_{\mathrm{tan}} \:=\:\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$$$\: \\ $$$$\:\therefore \\ $$$$\:\mathrm{y}−\mathrm{y}_{\mathrm{1}} \:=\:\mathrm{m}_{\mathrm{tan}} \left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right) \\ $$$$\:\mathrm{y}\:=\:\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{12}}\centerdot\left(\mathrm{z}\:−\:\frac{\mathrm{3}}{\mathrm{2}}\right)\:+\:\sqrt{\mathrm{3}} \\ $$$$\: \\ $$$$\:\mathrm{Thus},\:\mathrm{we}\:\mathrm{can}\:\mathrm{now}\:\mathrm{see}\:\mathrm{all}\:\mathrm{of}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{12}}\centerdot\left(\mathrm{z}\:−\:\frac{\mathrm{3}}{\mathrm{2}}\right)\:+\:\sqrt{\mathrm{3}}\:=\:\sqrt{\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{z}}{\mathrm{4}}} \\ $$$$\:\mathrm{z}\:=\:\frac{\mathrm{25}}{\mathrm{48}} \\ $$$$\: \\ $$$$\:\mathrm{Therefore},\:\mathrm{we}\:\mathrm{now}\:\mathrm{have}: \\ $$$$\:\mathrm{y}^{\mathrm{2}} \:=\:\left(\frac{\mathrm{25}}{\mathrm{48}}\right)^{\mathrm{3}} \:−\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\mathrm{25}}{\mathrm{48}}\right) \\ $$$$\:\mathrm{y}\:=\:\pm\:\frac{\mathrm{35}\sqrt{\mathrm{3}}}{\mathrm{576}} \\ $$$$\: \\ $$$$\:\therefore \\ $$$$\:\left(\mathrm{y},\:\mathrm{z}\right)\:=\:\left(\pm\:\frac{\mathrm{35}\sqrt{\mathrm{3}}}{\mathrm{576}},\:\frac{\mathrm{25}}{\mathrm{48}}\right)\:,\:\left(\pm\sqrt{\mathrm{3}}\:,\:\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\: \\ $$$$\:\mathrm{By}\:\mathrm{substituting}\:\mathrm{back},\:\mathrm{we}\:\mathrm{now}\:\mathrm{have}: \\ $$$$\:\mathrm{3m}^{\mathrm{2}} \:=\:\left(\pm\:\frac{\mathrm{35}\sqrt{\mathrm{3}}}{\mathrm{576}}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{m}\:=\:\pm\:\frac{\mathrm{35}}{\mathrm{576}} \\ $$$$\: \\ $$$$\:\mathrm{and}\: \\ $$$$\: \\ $$$$\:\frac{\mathrm{25}}{\mathrm{48}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{n} \\ $$$$\:\mathrm{n}\:=\:\frac{\mathrm{1}}{\mathrm{48}} \\ $$$$\: \\ $$$$\:\therefore \\ $$$$\:\left(\mathrm{m},\:\mathrm{n}\right)\:=\:\left(\pm\mathrm{1},\:\mathrm{1}\right),\:\left(\pm\:\frac{\mathrm{35}}{\mathrm{576}}\:,\:\frac{\mathrm{1}}{\mathrm{48}}\right) \\ $$$$\: \\ $$$$\: \\ $$$$\:\mathrm{By}\:\mathrm{graphing}\:\mathrm{the}\:\mathrm{elliptic}\:\mathrm{curve}\:\mathrm{and}\:\mathrm{its}\:\mathrm{tangent}\:\mathrm{line}: \\ $$$$\: \\ $$

Commented by Rasheed.Sindhi last updated on 26/Jan/22

n,m must be natural!

$$\mathrm{n},\mathrm{m}\:{must}\:{be}\:{natural}! \\ $$

Answered by Sheenaynay last updated on 26/Jan/22

((n×(n+1)×(2n+1))/6)=m^2 n(n+1)(2n+1)=m^2 ×6 n=24 ∧ m=70

$$\frac{\mathrm{n}×\left(\mathrm{n}+\mathrm{1}\right)×\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}=\mathrm{m}^{\mathrm{2}} \\ $$$$\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)=\mathrm{m}^{\mathrm{2}} ×\mathrm{6} \\ $$$$\mathrm{n}=\mathrm{24}\:\:\:\:\:\:\wedge\:\:\:\:\:\:\:\mathrm{m}=\mathrm{70} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com