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Question-165098

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

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