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

$p r o v e t h a t a r e a x a c t l y 1729 p o s i t i v e$ $i n t e g e r s o l u t i o n s t o t h e b e l o w e q u a t i o n$ $4 x^{4} + 3 y^{3} + 2 z^{2} + t = 4311$
Commented by Rasheed.Sindhi last updated on 27/Apr/21

$A T r y$ $^{∙} x ⩾ 1$ $M a x i m u m v a l u e o f x :$ $4 x^{4} + 3 {(1)}^{3} + 2 {(1)}^{2} + (1) = 4311$ $x^{4} = \frac{4311 - 6}{4} \Rightarrow x = 5.7276 . .$ $x ⩽ 5$ $p o s s i b l e v a l u e s o f x : 1, 2, 3, 4, 5$ $^{∙} y ⩾ 1$ $M a x i m u m v a l u e o f y :$ $4 {(1)}^{4} + 3 y^{3} + 2 {(1)}^{2} + (1) = 4311$ $y^{3} = \frac{4311 - 4 - 2 - 1}{3}$ $y = 11.278$ $y ⩽ 11$ $p o s s i b l e v a l u e s o f y : 1, 2, 3, . . ., 11$ $^{∙} z ⩾ 1$ $M a x i m u m v a l u e o f z :$ $4 {(1)}^{4} + 3 {(1)}^{3} + 2 z^{2} + (1) = 4311$ $z^{2} = \frac{4311 - 4 - 3 - 1}{2}$ $z = 46.384$ $z ⩽ 46$ $p o s s i b l e v a l u e s o f z : 1, 2, 3, . . ., 46$ $^{∙} t ⩾ 1$ $M a x i m u m v a l u e o f t :$ $4 {(1)}^{4} + 3 {(1)}^{3} + 2 {(1)}^{2} + t = 4311$ $t = 4311 - 4 - 3 - 2 = 4302$ $t ⩽ 4302$ $p o s s i b l e v a l u e s o f t : 1, 2, 3, . . ., 4302$ $C o n t i n u e$
Commented by mathsuji last updated on 28/Apr/21

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