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Question Number 144439 by mathdanisur last updated on 25/Jun/21

Commented by Rasheed.Sindhi last updated on 25/Jun/21

$g c d (m^{2} - m + 1, m + 1) = 1 ?$
Commented by mathdanisur last updated on 25/Jun/21

$s o r r y S i r, y e s + 1$
Commented by mathmax by abdo last updated on 25/Jun/21

$determine u and v / u (m^{2} - m + 1) + v (m + 1) = 1$ $u = k and v = α m + β . . . .$
Commented by Rasheed.Sindhi last updated on 26/Jun/21

$T {h e r e}^{'} s a s o l u t i o n w h i c h s a t i s f i e s$ $m^{4} + m - {p n}^{2} = 0, b u t w h i c h {d o e s n}^{'} t$ $f u l f i l l t h e c o d i t i o n$ $g c d (m^{2} - m + 1, m + 1) = 1$ $S o l u t i o n : m = p = 2, n = 3 :$ $2^{4} + 2 - 2 (3^{2}) = 0$ $16 + 2 - 18 = 0$ $0 = 0$ $b u t g c d (2^{2} - 2 + 1, 2 + 1) \overset{?}{=} 1$ $g c d (3, 3) = 3 \neq 1$ $H o w e v e r i t f u l f i l l s$ $g c d (m^{2} - m - 1, m + 1) = 1$ $g c d (2^{2} - 2 - 1, 2 + 1) \overset{?}{=} 1$ $g c d (1, 3) = 1$
	Answered by Rasheed.Sindhi last updated on 26/Jun/21
	$∀ ⊥┌⋌ (Not a perfect solution) { ((m^4 +m−pn^2 =0)),((m,n∈N,p∈P, gcd(m^2 −m+1,m+1)=1)) :} m(m+1)(m^2 −m+1)=pn^2 SOME CASES: determinant (((m=1 ∧ (m+1)(m^2 −m+1)=pn^2 )),((m=p ∧ (m+1)(m^2 −m+1)=n^2 )),((m=n ∧ (m+1)(m^2 −m+1)=pn)),((m=n^2 ∧ (m+1)(m^2 −m+1)=p)),((m=pn ∧ (m+1)(m^2 −m+1)=n))) C-1: m ∣ p⇒m=1∨m=p ^• m=1 ∧ (m+1)(m^2 −m+1)=pn^2 (1+1)(1^2 −1+1)=pn^2 pn^2 =2⇒p=2 ∧ n^2 =1 m=n=1,p=2 gcd(m^2 −m+1,m+1)=1 gcd(1^2 −1+1,1+1)=1 gcd(1,2)=1 ^• m=p ∧ (m+1)(m^2 −m+1)=n^2 (p+1)(p^2 −p+1)=n^2 p^3 +1=n^2 (We′ve to find such primes p for which p^3 +1 is perfect square) For p=2, p^3 +1=9(perfect square) ∴m=p=2,n=3 But this doesn′t fulfill the condition gcd(m^2 −m+1,m+1) =gcd(2^2 −2+1,2+1)=gcd(3,3)=3≠1 ^• p^3 +1=n^2 (p+1)(p^2 −p+1)=n^2 p+1=p^2 −p+1=n p^2 −2p=0 p−2=0 p=2 ⇒n=p+1=2+1=3 m=p=2,n=3 (same result) ^• m=n ∧ (m+1)(m^2 −m+1)=pn (n+1)(n^2 −n+1)=pn n^3 +1=pn n^2 +(1/n)=p Clearly p is whole number(prime) so n=1⇒m=1 so p=2 (gcd(1^2 −1+1,1+1)=gcd(1,2)=1) ^• m=n^2 ∧ (m+1)(m^2 −m+1)=p (n^2 +1)(n^2 −n+1)=p ⇒ n^2 +1=1 ∧ n^2 −n+1=p......(i) OR n^2 +1=p ∧ n^2 −n+1=1.....(ii) (i) n=0 ∧ p=1( false ∵ p∈P) (ii) n^2 −n+1=1⇒n(n−1)=0 ⇒n=0 ∣ n=1 n^2 +1=p 0^2 +1=p ∣ 1^2 +1=p p=1 ∣ p=2 (false) ∣ n=1⇒m=n^2 =1^2 =1 ∣ gcd(1^2 −1+1,1+1)=1 ....... .....$
	$\forall ⊥ ⌜ ⋌$ $(N o t a p e r f e c t s o l u t i o n)$ ${\begin{cases} m^{4} + m - {p n}^{2} = 0 \\ m, n \in N, p \in P, g c d (m^{2} - m + 1, m + 1) = 1 \end{cases}$ $m (m + 1) (m^{2} - m + 1) = {p n}^{2}$ $SOME CASES :$ $\begin{matrix} m = 1 \land (m + 1) (m^{2} - m + 1) = {p n}^{2} \\ m = p \land (m + 1) (m^{2} - m + 1) = n^{2} \\ m = n \land (m + 1) (m^{2} - m + 1) = p n \\ m = n^{2} \land (m + 1) (m^{2} - m + 1) = p \\ m = p n \land (m + 1) (m^{2} - m + 1) = n \end{matrix}$ $C - 1 : m ∣ p \Rightarrow m = 1 \lor m = p$ $^{∙} m = 1 \land (m + 1) (m^{2} - m + 1) = {p n}^{2}$ $(1 + 1) (1^{2} - 1 + 1) = {p n}^{2}$ ${p n}^{2} = 2 \Rightarrow p = 2 \land n^{2} = 1$ $m = n = 1, p = 2$ $g c d (m^{2} - m + 1, m + 1) = 1$ $g c d (1^{2} - 1 + 1, 1 + 1) = 1$ $g c d (1, 2) = 1$ $^{∙} m = p \land (m + 1) (m^{2} - m + 1) = n^{2}$ $(p + 1) (p^{2} - p + 1) = n^{2}$ $p^{3} + 1 = n^{2}$ $({W e}^{'} v e t o f i n d s u c h p r i m e s p$ $f o r w h i c h p^{3} + 1 i s p e r f e c t s q u a r e)$ $F o r p = 2, p^{3} + 1 = 9 (p e r f e c t s q u a r e)$ $∴ m = p = 2, n = 3$ $B u t t h i s {d o e s n}^{'} t f u l f i l l t h e c o n d i t i o n$ $g c d (m^{2} - m + 1, m + 1)$ $= g c d (2^{2} - 2 + 1, 2 + 1) = g c d (3, 3) = 3 \neq 1$ $^{∙} p^{3} + 1 = n^{2}$ $(p + 1) (p^{2} - p + 1) = n^{2}$ $p + 1 = p^{2} - p + 1 = n$ $p^{2} - 2 p = 0$ $p - 2 = 0$ $p = 2 \Rightarrow n = p + 1 = 2 + 1 = 3$ $m = p = 2, n = 3 (s a m e r e s u l t)$ $^{∙} m = n \land (m + 1) (m^{2} - m + 1) = p n$ $(n + 1) (n^{2} - n + 1) = p n$ $n^{3} + 1 = p n$ $n^{2} + \frac{1}{n} = p$ $C l e a r l y p i s w h o l e n u m b e r (p r i m e)$ $s o n = 1 \Rightarrow m = 1$ $s o p = 2$ $(g c d (1^{2} - 1 + 1, 1 + 1) = g c d (1, 2) = 1)$ $^{∙} m = n^{2} \land (m + 1) (m^{2} - m + 1) = p$ $(n^{2} + 1) (n^{2} - n + 1) = p$ $\Rightarrow n^{2} + 1 = 1 \land n^{2} - n + 1 = p . . . . . . (i)$ $OR$ $n^{2} + 1 = p \land n^{2} - n + 1 = 1 . . . . . (i i)$ $(i) n = 0 \land p = 1 (f a l s e ∵ p \in P)$ $(i i) n^{2} - n + 1 = 1 \Rightarrow n (n - 1) = 0$ $\Rightarrow n = 0 ∣ n = 1$ $n^{2} + 1 = p$ $0^{2} + 1 = p ∣ 1^{2} + 1 = p$ $p = 1 ∣ p = 2$ $(f a l s e) ∣ n = 1 \Rightarrow m = n^{2} = 1^{2} = 1$ $∣ g c d (1^{2} - 1 + 1, 1 + 1) = 1$ $. . . . . . .$ $. . . . .$
	Commented by mathdanisur last updated on 26/Jun/21

	$H o w S i r . .$
	Commented by Rasheed.Sindhi last updated on 26/Jun/21

	$O n e d i t i n g, a g r e a t p a r t o f m y a n s w e r h a s$ $b e e n d i s a p p e a r e d!!! {D o n}^{'} t k n o w$ $w h y ? P e r h a p s b e c a u s e m y a n s w e r$ $w a s t o o l e n g t h y!$
	Commented by mathdanisur last updated on 26/Jun/21

	${I t}^{'} s a p i t y s o m e t i m e s, t h a n k y o u S i r$
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