Solve over integer a^3 +2a^2 +3a+4=b!+c!


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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

	$a^{3} + {2 a}^{2} + 3 a + 4 = b! + c!$ $a^{3} + {2 a}^{2} + 3 a + 4 - (b! + c!) = 0$ $T h e e q u a t i o n i s c u b i c$ $∴ a has at most 3 integral values$ $f o r p a r t i c u l a r v a l u e o f b & c .$ $Let α, β & γ a r e t h e r o o t s o f t h e a b o v e e q u a t i o n .$ $α + β + γ = - 2, α β + β γ + γ α = 3,$ $α β γ = b! + c! - 4$ $. . . . .$ $. . . .$
	Commented by dragan91 last updated on 16/Jul/22
	$a^3 +2a^2 +3a+4=b!+c! (a+1)(a^2 +a+2)=b!+c!−2 since b!+c!−2≥0 ∀b,c∈N ⇒(a+1)(a^2 +a+2)≥0⇒a≥−1 a^3 +2a^2 +3a+4 is never divisible by 3 then for 3≤b≤c and b!+c!=3 is no solution a^3 +2a^2 +3a+4 could have solutions for 1+c! (c≠2) or 2+c! (c≠1) We have to consider cases: I)for fixed b!=1⇒b=0=1 c≠2 a)assume a≤c⇒c!≡0(mod a) a^3 +2a^2 +3a+4=1+c! a^3 +2a^2 +3a+3=c! /:a a^2 +2a+3+^ (3/a)=((c!)/a) LHS is integer for a={(−1,1,2,3)} ⇒(a,b,c)={(−1,0,0),(−1,1,0),(−1,0,1) (−1,1,1),(2,4,2),(2,2,4)} b)assume c<a c!<a! a^3 +2a^2 +3a+3<a! its true for a≥6 for a≥6 let a=6+k (k≥0) (6+k)^3 +2(6+k)^2 +3(6+k)+3=c! k^3 +20k^2 +35k+309=c! LHS is not divisible by 2 for any inter ⇒ c=0 or c=1 ⇒k=−7 is no solution (k≥0) c)a=0 no solution II)For fixed b!=2⇒b=2 a^3 +2a^2 +3a+2=c! a)c≥a ⇒c!≡0(mod a) a^3 +2a^2 +3a+2=c!/:a a^2 +2a+3a+(2/a)=((c!)/a) LHS is integer for a={(−1,1,2)} which is already checked solutions b)c<a a^3 +2a^2 +3a+2<a! its true for a≥6⇒a=6+p p≥0 c<6+p (6+p)^3 +2(6+p)^2 +3(6+p)+2=c! p^3 +20p^2 +135p+308=c! p^3 +3p^2 ∙7+3p∙49+343−(p^2 +12p+35)=c! (p+7)^3 −(p+7)(p+5)=c! p+7=t t^3 −t(t−2)=c! Since c<p+6⇒c<t−1<t for t>1 c≢0(mod t)⇒c!≢0(mod t) t^3 −t(t−2)≡0(mod t) its only possible solution for t=1 c!=1^3 −1^2 +2⇒c=2 Its contradiction since c<t−1 c)a=0⇒c=2 (a,b,c)={(0,2,2)} All solutions (a,b,c)={(−1,0,0),(−1,0,1),(−1,1,0), (−1,1,1),(0,2,2),(2,2,4),(2,4,2)}$
	$\underset{―}{a^{3} + {2 a}^{2} + 3 a + 4 = b! + c!}$ $(a + 1) (a^{2} + a + 2) = b! + c! - 2$ $since b! + c! - 2 ⩾ 0 \forall b, c \in N$ $\Rightarrow (a + 1) (a^{2} + a + 2) ⩾ 0 \Rightarrow a ⩾ - 1$ $a^{3} + {2 a}^{2} + 3 a + 4 is never divisible by 3$ $then for 3 ⩽ b ⩽ c and b! + c! = 3$ $is no solution$ $a^{3} + {2 a}^{2} + 3 a + 4 could have solutions$ $for 1 + c! (c \neq 2) or 2 + c! (c \neq 1)$ $We have to consider cases :$ $I) for fixed b! = 1 \Rightarrow b = 0 = 1 c \neq 2$ $a) assume a ⩽ c \Rightarrow c! \equiv 0 (\mod a)$ $a^{3} + {2 a}^{2} + 3 a + 4 = 1 + c!$ $a^{3} + {2 a}^{2} + 3 a + 3 = c! / : a$ $a^{2} + 2 a + 3 +^{} \frac{3}{a} = \frac{c!}{a}$ $LHS is integer for a = {(- 1, 1, 2, 3)}$ $\Rightarrow (a, b, c) = {(- 1, 0, 0), (- 1, 1, 0), (- 1, 0, 1)$ $(- 1, 1, 1), (2, 4, 2), (2, 2, 4)}$ $b) assume c < a$ $c! < a!$ $a^{3} + {2 a}^{2} + 3 a + 3 < a!$ $its true for a ⩾ 6$ $for a ⩾ 6 let a = 6 + k (k ⩾ 0)$ ${(6 + k)}^{3} + 2 {(6 + k)}^{2} + 3 (6 + k) + 3 = c!$ $k^{3} + {20 k}^{2} + 35 k + 309 = c!$ $LHS is not divisible by 2$ $for any inter \Rightarrow c = 0 or c = 1$ $\Rightarrow k = - 7 is no solution (k ⩾ 0)$ $c) a = 0 no solution$ $II) For fixed b! = 2 \Rightarrow b = 2$ $a^{3} + {2 a}^{2} + 3 a + 2 = c!$ $a) c ⩾ a \Rightarrow c! \equiv 0 (\mod a)$ $a^{3} + {2 a}^{2} + 3 a + 2 = c! / : a$ $a^{2} + 2 a + 3 a + \frac{2}{a} = \frac{c!}{a}$ $LHS is integer for a = {(- 1, 1, 2)} which is$ $already checked solutions$ $b) c < a$ $a^{3} + {2 a}^{2} + 3 a + 2 < a!$ $its true for a ⩾ 6 \Rightarrow a = 6 + p p ⩾ 0$ $c < 6 + p$ ${(6 + p)}^{3} + 2 {(6 + p)}^{2} + 3 (6 + p) + 2 = c!$ $p^{3} + {20 p}^{2} + 135 p + 308 = c!$ $p^{3} + {3 p}^{2} \cdot 7 + 3 p \cdot 49 + 343 - (p^{2} + 12 p + 35) = c!$ ${(p + 7)}^{3} - (p + 7) (p + 5) = c!$ $p + 7 = t$ $t^{3} - t (t - 2) = c!$ $Since c < p + 6 \Rightarrow c < t - 1 < t$ $for t > 1 c ≢ 0 (\mod t) \Rightarrow c! ≢ 0 (\mod t)$ $t^{3} - t (t - 2) \equiv 0 (\mod t)$ $its only possible solution for t = 1$ $c! = 1^{3} - 1^{2} + 2 \Rightarrow c = 2$ $Its contradiction since c < t - 1$ $c) a = 0 \Rightarrow c = 2$ $(a, b, c) = {(0, 2, 2)}$ $\underset{―}{}$ $All solutions$ $(a, b, c) = {(- 1, 0, 0), (- 1, 0, 1), (- 1, 1, 0),$ $(- 1, 1, 1), (0, 2, 2), (2, 2, 4), (2, 4, 2)}$
	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

	$α + β + γ = - 2 \land α β + β γ + γ α = 3$ $\Rightarrow α β γ m a y h a v e m u l t i p l e v a l u e s .$ $S o m e o t h e r s o l u t i o n s o f t h e e q u a t i o n :$ $(a, b, c) = (- 1, 0, 1), (- 1, 1, 0), (- 1, 1, 1),$ $(2, 2, 4), (2, 4, 2), . . .$
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