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prove-that-are-axactly-1729-positive-integer-solutions-to-the-below-equation-4x-4-3y-3-2z-2-t-4311-

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Question Number 139319 by mathsuji last updated on 25/Apr/21
prove that are axactly 1729 positive integer solutions to the below equation 4x^4 +3y^3 +2z^2 +t=4311
$${prove}\:{that}\:{are}\:{axactly}\:\mathrm{1729}\:{positive} \\ $$$${integer}\:{solutions}\:{to}\:{the}\:{below}\:{equation} \\ $$$$\mathrm{4}{x}^{\mathrm{4}} +\mathrm{3}{y}^{\mathrm{3}} +\mathrm{2}{z}^{\mathrm{2}} +{t}=\mathrm{4311} \\ $$
Commented by Rasheed.Sindhi last updated on 27/Apr/21
A Try ^• x≥1 Maximum value of x: 4x^4 +3(1)^3 +2(1)^2 +(1)=4311 x^4 =((4311−6)/4)⇒x=5.7276.. x≤5 possible values of x:1,2,3,4,5 ^• y≥1 Maximum value of y: 4(1)^4 +3y^3 +2(1)^2 +(1)=4311 y^3 =((4311−4−2−1)/3) y=11.278 y≤11 possible values of y:1,2,3,...,11 ^• z≥1 Maximum value of z: 4(1)^4 +3(1)^3 +2z^2 +(1)=4311 z^2 =((4311−4−3−1)/2) z=46.384 z≤46 possible values of z:1,2,3,...,46 ^• t≥1 Maximum value of t: 4(1)^4 +3(1)^3 +2(1)^2 +t=4311 t=4311−4−3−2=4302 t≤4302 possible values of t:1,2,3,...,4302 Continue
$${A}\:{Try} \\ $$$$\:^{\bullet} {x}\geqslant\mathrm{1} \\ $$$${Maximum}\:{value}\:{of}\:{x}: \\ $$$$\mathrm{4}{x}^{\mathrm{4}} +\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)=\mathrm{4311} \\ $$$${x}^{\mathrm{4}} =\frac{\mathrm{4311}−\mathrm{6}}{\mathrm{4}}\Rightarrow{x}=\mathrm{5}.\mathrm{7276}.. \\ $$$${x}\leqslant\mathrm{5} \\ $$$${possible}\:{values}\:{of}\:{x}:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5} \\ $$$$\:^{\bullet} {y}\geqslant\mathrm{1} \\ $$$${Maximum}\:{value}\:{of}\:{y}: \\ $$$$\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{4}} +\mathrm{3}{y}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)=\mathrm{4311} \\ $$$$\:\:\:\:\:\:{y}^{\mathrm{3}} =\frac{\mathrm{4311}−\mathrm{4}−\mathrm{2}−\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:{y}=\mathrm{11}.\mathrm{278} \\ $$$$\:\:\:\:{y}\leqslant\mathrm{11} \\ $$$${possible}\:{values}\:{of}\:{y}:\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{11} \\ $$$$\:^{\bullet} {z}\geqslant\mathrm{1} \\ $$$$\:\:\:{Maximum}\:{value}\:{of}\:{z}: \\ $$$$\:\:\:\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{4}} +\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}{z}^{\mathrm{2}} +\left(\mathrm{1}\right)=\mathrm{4311} \\ $$$$\:\:\:\:\:{z}^{\mathrm{2}} =\frac{\mathrm{4311}−\mathrm{4}−\mathrm{3}−\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{z}=\mathrm{46}.\mathrm{384} \\ $$$${z}\leqslant\mathrm{46} \\ $$$${possible}\:{values}\:{of}\:{z}:\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{46} \\ $$$$\:^{\bullet} {t}\geqslant\mathrm{1} \\ $$$$\:\:\:{Maximum}\:{value}\:{of}\:{t}: \\ $$$$\:\:\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{4}} +\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} +{t}=\mathrm{4311} \\ $$$$\:\:\:\:\:{t}=\mathrm{4311}−\mathrm{4}−\mathrm{3}−\mathrm{2}=\mathrm{4302} \\ $$$$\:{t}\leqslant\mathrm{4302} \\ $$$${possible}\:{values}\:{of}\:{t}:\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{4302} \\ $$$${Continue} \\ $$$$ \\ $$
Commented by mathsuji last updated on 28/Apr/21
cool thankyou sir
$${cool}\:{thankyou}\:{sir} \\ $$

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