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

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, \dots, 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, \dots, 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, \dots, 4302

C o n t i n u e

Commented by mathsuji last updated on 28/Apr/21

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