Solve-over-integer-a-3-2a-2-3a-4-b-c-

Question Number 173556 by dragan91 last updated on 13/Jul/22

Solve over integer

a^{3} + {2 a}^{2} + 3 a + 4 = b! + c!

Answered by Rasheed.Sindhi last updated on 14/Jul/22

a^{3} + {2 a}^{2} + 3 a + 4 = b! + c!

a^{3} + {2 a}^{2} + 3 a + 4 - (b! + c!) = 0

T h e e q u a t i o n i s c u b i c

∴ a has at most 3 integral values

f o r p a r t i c u l a r v a l u e o f b & c .

Let α, β & γ a r e t h e r o o t s o f t h e a b o v e e q u a t i o n .

α + β + γ = - 2, α β + β γ + γ α = 3,

α β γ = b! + c! - 4

\dots . .

\dots .

Commented by dragan91 last updated on 16/Jul/22

\underset{―}{a^{3} + {2 a}^{2} + 3 a + 4 = b! + c!}

(a + 1) (a^{2} + a + 2) = b! + c! - 2

since b! + c! - 2 ⩾ 0 \forall b, c \in N

\Rightarrow (a + 1) (a^{2} + a + 2) ⩾ 0 \Rightarrow a ⩾ - 1

a^{3} + {2 a}^{2} + 3 a + 4 is never divisible by 3

then for 3 ⩽ b ⩽ c and b! + c! = 3

is no solution

a^{3} + {2 a}^{2} + 3 a + 4 could have solutions

for 1 + c! (c \neq 2) or 2 + c! (c \neq 1)

We have to consider cases :

I) for fixed b! = 1 \Rightarrow b = 0 = 1 c \neq 2

a) assume a ⩽ c \Rightarrow c! \equiv 0 (\mod a)

a^{3} + {2 a}^{2} + 3 a + 4 = 1 + c!

a^{3} + {2 a}^{2} + 3 a + 3 = c! / : a

a^{2} + 2 a + 3 +^{} \frac{3}{a} = \frac{c!}{a}

LHS is integer for a = {(- 1, 1, 2, 3)}

\Rightarrow (a, b, c) = {(- 1, 0, 0), (- 1, 1, 0), (- 1, 0, 1)

(- 1, 1, 1), (2, 4, 2), (2, 2, 4)}

b) assume c < a

c! < a!

a^{3} + {2 a}^{2} + 3 a + 3 < a!

its true for a ⩾ 6

for a ⩾ 6 let a = 6 + k (k ⩾ 0)

{(6 + k)}^{3} + 2 {(6 + k)}^{2} + 3 (6 + k) + 3 = c!

k^{3} + {20 k}^{2} + 35 k + 309 = c!

LHS is not divisible by 2

for any inter \Rightarrow c = 0 or c = 1

\Rightarrow k = - 7 is no solution (k ⩾ 0)

c) a = 0 no solution

II) For fixed b! = 2 \Rightarrow b = 2

a^{3} + {2 a}^{2} + 3 a + 2 = c!

a) c ⩾ a \Rightarrow c! \equiv 0 (\mod a)

a^{3} + {2 a}^{2} + 3 a + 2 = c! / : a

a^{2} + 2 a + 3 a + \frac{2}{a} = \frac{c!}{a}

LHS is integer for a = {(- 1, 1, 2)} which is

already checked solutions

b) c < a

a^{3} + {2 a}^{2} + 3 a + 2 < a!

its true for a ⩾ 6 \Rightarrow a = 6 + p p ⩾ 0

c < 6 + p

{(6 + p)}^{3} + 2 {(6 + p)}^{2} + 3 (6 + p) + 2 = c!

p^{3} + {20 p}^{2} + 135 p + 308 = c!

p^{3} + {3 p}^{2} \cdot 7 + 3 p \cdot 49 + 343 - (p^{2} + 12 p + 35) = c!

{(p + 7)}^{3} - (p + 7) (p + 5) = c!

p + 7 = t

t^{3} - t (t - 2) = c!

Since c < p + 6 \Rightarrow c < t - 1 < t

for t > 1 c ≢ 0 (\mod t) \Rightarrow c! ≢ 0 (\mod t)

t^{3} - t (t - 2) \equiv 0 (\mod t)

its only possible solution for t = 1

c! = 1^{3} - 1^{2} + 2 \Rightarrow c = 2

Its contradiction since c < t - 1

c) a = 0 \Rightarrow c = 2

(a, b, c) = {(0, 2, 2)}

\underset{―}{}

All solutions

(a, b, c) = {(- 1, 0, 0), (- 1, 0, 1), (- 1, 1, 0),

(- 1, 1, 1), (0, 2, 2), (2, 2, 4), (2, 4, 2)}

Commented by dragan91 last updated on 15/Jul/22

yes . but equation has more solution wuth this conditions . α β γ = b! + c! - 4

has infinite solutions .

Commented by Rasheed.Sindhi last updated on 16/Jul/22

α + β + γ = - 2 \land α β + β γ + γ α = 3

\Rightarrow α β γ m a y h a v e m u l t i p l e v a l u e s .

S o m e o t h e r s o l u t i o n s o f t h e e q u a t i o n :

(a, b, c) = (- 1, 0, 1), (- 1, 1, 0), (- 1, 1, 1),

(2, 2, 4), (2, 4, 2), \dots