w,x,y,z are digits in respective base system and a,b are bases. Find out an example/examples which satisfy the following wxyz_a +wxyz_b =wxyz


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Question Number 7249 by Rasheed Soomro last updated on 19/Aug/16

$w, x, y, z a r e d i g i t s i n r e s p e c t i v e b a s e s y s t e m a n d$ $a, b a r e b a s e s .$ $F i n d o u t a n e x a m p l e / e x a m p l e s w h i c h s a t i s f y$ $t h e f o l l o w i n g$ ${w x y z}_{a} + {w x y z}_{b} = {w x y z}_{a + b}$
Commented by Yozzia last updated on 19/Aug/16
$2^1 +2^1 =(2+2)^1 =4^1 {1×2+0×2^0 }+{1×2+0×2^0 }=1×4+0×4^0 or (10)_2 +(10)_2 =(10)_4$
$2^{1} + 2^{1} = {(2 + 2)}^{1} = 4^{1}$ ${1 \times 2 + 0 \times 2^{0}} + {1 \times 2 + 0 \times 2^{0}} = 1 \times 4 + 0 \times 4^{0}$ $o r {(10)}_{2} + {(10)}_{2} = {(10)}_{4}$
Commented by Rasheed Soomro last updated on 19/Aug/16

$F o u r d i g i t n u m b e r i s r e q u i r e d .$
Commented by Yozzia last updated on 19/Aug/16
$wxyz_a +wxyz_b =wxyz_(a+b) (∗) Suppose wxyz is an integer identical on both sides of (∗) e.g 1234_a +1234_b =1234_(a+b) for some given a,b∈N. [wa^3 +xa^2 +ya+z]+[wb^3 +xb^2 +yb+z]=w(a+b)^3 +x(a+b)^2 +y(a+b)+z w{a^3 +b^3 −(a+b)^3 }+x{a^2 +b^2 −(a+b)^2 }+y(a+b−a−b)+2z−z=0 w{−3a^2 b−3ab^2 }+x{−2ab}+z=0 w{3a^2 b+3ab^2 }+2xab−z=0 The form of the above equation is independent of the value of y∈N. z=ab(2x+3w(a+b)) w,x∈Z^≥ , a,b∈N−{1} and ab∣z. If x>0 or w>0 then 2x+3w(a+b)>1⇒z>ab>a and z>b. However, 0≤z≤min(a−1,b−1)<min(a,b) for z in wxyz_a and wxyz_b , a contradiction. Hence, we must have x=w=0 ⇒ z=0. ∴ we get 00y0_a +00y0_b =00y0_(a+b) or y0_a +y0_b =y0_(a+b) or ya+yb=y(a+b) for any y∈N ,0≤y<a,y<b when a and b are known bases. No four digit integer wxyz satisfies wxyz_a +wxyz_b =wxyz_(a+b) .$
${w x y z}_{a} + {w x y z}_{b} = {w x y z}_{a + b} ()$ $S u p p o s e w x y z i s a n i n t e g e r i d e n t i c a l o n b o t h s i d e s o f ()$ $e . g 1234_{a} + 1234_{b} = 1234_{a + b}$ $f o r s o m e g i v e n a, b \in N .$ $[{w a}^{3} + {x a}^{2} + y a + z] + [{w b}^{3} + {x b}^{2} + y b + z] = w {(a + b)}^{3} + x {(a + b)}^{2} + y (a + b) + z$ $w {a^{3} + b^{3} - {(a + b)}^{3}} + x {a^{2} + b^{2} - {(a + b)}^{2}} + y (a + b - a - b) + 2 z - z = 0$ $w {- 3 a^{2} b - 3 {a b}^{2}} + x {- 2 a b} + z = 0$ $w {3 a^{2} b + 3 {a b}^{2}} + 2 x a b - z = 0$ $T h e f o r m o f t h e a b o v e e q u a t i o n i s$ $i n d e p e n d e n t o f t h e v a l u e o f y \in N .$ $z = a b (2 x + 3 w (a + b))$ $w, x \in Z^{⩾}, a, b \in N - {1} a n d a b ∣ z . I f x > 0 o r w > 0$ $t h e n 2 x + 3 w (a + b) > 1 \Rightarrow z > a b > a a n d z > b .$ $H o w e v e r, 0 ⩽ z ⩽ m i n (a - 1, b - 1) < m i n (a, b)$ $f o r z i n {w x y z}_{a} a n d {w x y z}_{b}, a c o n t r a d i c t i o n .$ $H e n c e, w e m u s t h a v e x = w = 0 \Rightarrow z = 0 .$ $∴ w e g e t 00 y 0_{a} + 00 y 0_{b} = 00 y 0_{a + b} o r y 0_{a} + y 0_{b} = y 0_{a + b}$ $o r y a + y b = y (a + b) f o r a n y y \in N, 0 ⩽ y < a, y < b$ $w h e n a a n d b a r e k n o w n b a s e s .$ $N o f o u r d i g i t i n t e g e r w x y z s a t i s f i e s$ ${w x y z}_{a} + {w x y z}_{b} = {w x y z}_{a + b} .$
Commented by Rasheed Soomro last updated on 20/Aug/16

$V . N i c E!$ $W h a t i s m e a n t b y : w, x \in Z^{⩾} ?$
Commented by Yozzia last updated on 20/Aug/16
$Z^≥ ={0,1,2,3,4,...} w,x∈Z^≥ ≡x,w are non−negative integers.$
$Z^{⩾} = {0, 1, 2, 3, 4, . . .}$ $w, x \in Z^{⩾} \equiv x, w a r e n o n - n e g a t i v e i n t e g e r s .$
Commented by Rasheed Soomro last updated on 20/Aug/16

$T h α n k S!$
Commented by Yozzia last updated on 20/Aug/16
$Let q be a number of digit length n≥3, and c,b∈[N−{1}]. Write k_m for the number k corresponding to base m≥2. Suppose that q_c +q_b =q_(c+b) . Then, no q exists satisfying this equation for n≥3. −−−−−−−−−−−−−−−−−−−−−−−−−−−−−− PROOF: Looking at the equation q_c +q_b =q_(c+b) , q has a digit representation of the form q=a_0 ^� a_1 ^� a_2 ^� a_3 ^� ...a_(n−1) ^� where 0≤a_i ≤min(c−1,b−1) for i=1 , 2 , ... , n−1, and 1≤a_0 ≤min(c−1,b−1). Therefore q_c =Σ_(i=0) ^(n−1) a_i c^(n−1−i) , q_b =Σ_(i=1) ^(n−1) a_i b^(n−1−i) and q_(c+b) =Σ_(i=0) ^(n−1) a_i (b+c)^(n−1−i) . Σ_(i=0) ^(n−1) a_i c^(n−1−i) +Σ_(i=0) ^(n−1) a_i b^(n−1−i) =Σ_(i=0) ^(n−1) a_i (b+c)^(n−1−i) Σ_(i=0) ^(n−1) {a_i c^(n−1−i) +a_i b^(n−1−i) −a_i (b+c)^(n−1−i) }=0 Σ_(i=0) ^(n−1) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]=0 Σ_(i=0) ^(n−3) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]+a_(n−2) {c^(n−1−n+2) +b^(n−1−n+2) −(b+c)^(n−1−n+2) }+a_(n−1) {c^(n−1−n+1) +b^(n−1−n+1) −(b+c)^(n−1−n+1) }=0 Σ_(i=0) ^(n−3) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]+a_(n−2) {c+b−(b+c)}+a_(n−1) {1+1−1}=0 Σ_(i=0) ^(n−3) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]+a_(n−1) =0 a_(n−1) =Σ_(i=0) ^(n−3) [a_i {(b+c)^(n−1−i) −c^(n−1−i) −b^(n−1−i) }] According to the Binomial Theorem, (b+c)^(n−1−i) =Σ_(k=0) ^(n−1−i) (((n−1−i)),(k) ) b^(n−1−i−k) c^k or (b+c)^(n−1−i) =b^(n−1−i) +c^(n−1−i) +Σ_(k=1) ^(n−2−i) (((n−1−i)),(k) ) b^(n−1−i−k) c^k . ∴ (b+c)^(n−1−i) −b^(n−1−i) −c^(n−1−i) =Σ_(k=1) ^(n−2−i) (((n−1−i)),(k) ) b^(n−1−i−k) c^k . a_(n−1) =Σ_(i=0) ^(n−3) [a_i {Σ_(k=1) ^(n−2−i) (((n−1−i)),(k) ) b^(n−1−i−k) c^k }] a_(n−1) =a_0 {Σ_(k=1) ^(n−2) (((n−1)),(k) ) b^(n−1−k) c^k }+a_1 {Σ_(k=1) ^(n−3) (((n−2)),(k) ) b^(n−2−k) c^k } +a_2 {Σ_(k=1) ^(n−4) (((n−3)),(k) ) b^(n−3−k) c^k }+...+a_(n−4) {Σ_(k=1) ^2 ((3),(k) ) b^(3−k) c^k }+a_(n−3) {Σ_(k=1) ^1 ((2),(k) ) b^(2−k) c^k } Observe that the form of the above equation is indepedent of a_(n−2) ; so a_(n−2) is a free variable such that 0≤a_(n−2) ≤min(c−1,b−1). −−−−−−−−−−−−−−−−−−−−−−−−−−−−− Proposition: For any n≥3, bc divides Σ_(k=1) ^(n−2) (((n−1)),(k) ) b^(n−1−k) c^k . Proof (induction): Base case n=3: Σ_(k=1) ^(3−2) (((3−1)),(k) ) b^(3−1−k) c^k = ((2),(1) ) bc ⇒bc divides Σ_(k=1) ^(n−2) (((n−1)),(k) ) b^(n−1−k) c^k for n=3. Inductive step: Suppose bc divides Σ_(k=1) ^(j−2) (((j−1)),(k) ) b^(j−1−k) c^k for n=j. i.e Σ_(k=1) ^(j−2) (((j−1)),(k) ) b^(j−1−k) c^k =rbc for some r∈N. For n=j+1, we have that Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =Σ_(k=1) ^(j−1) { (((j−1)),(k) ) + (((j−1)),((k−1)) )}b^(j−k) c^k$
$L e t q b e a n u m b e r o f d i g i t l e n g t h n ⩾ 3,$ $a n d c, b \in [N - {1}] . W r i t e k_{m} f o r t h e$ $n u m b e r k c o r r e s p o n d i n g t o b a s e m ⩾ 2 .$ $S u p p o s e t h a t q_{c} + q_{b} = q_{c + b} .$ $T h e n, n o q e x i s t s s a t i s f y i n g t h i s e q u a t i o n$ $f o r n ⩾ 3 .$ $- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -$ $P R O O F :$ $L o o k i n g a t t h e e q u a t i o n q_{c} + q_{b} = q_{c + b},$ $q h a s a d i g i t r e p r e s e n t a t i o n o f t h e f o r m$ $q = {\bar{a}}_{0} {\bar{a}}_{1} {\bar{a}}_{2} {\bar{a}}_{3} . . . {\bar{a}}_{n - 1} w h e r e 0 ⩽ a_{i} ⩽ m i n (c - 1, b - 1)$ $f o r i = 1, 2, . . ., n - 1, a n d 1 ⩽ a_{0} ⩽ m i n (c - 1, b - 1) .$ $T h e r e f o r e q_{c} = \underset{i = 0}{\sum^{n - 1}} a_{i} c^{n - 1 - i}, q_{b} = \underset{i = 1}{\sum^{n - 1}} a_{i} b^{n - 1 - i}$ $a n d q_{c + b} = \underset{i = 0}{\sum^{n - 1}} a_{i} {(b + c)}^{n - 1 - i} .$ $\underset{i = 0}{\sum^{n - 1}} a_{i} c^{n - 1 - i} + \underset{i = 0}{\sum^{n - 1}} a_{i} b^{n - 1 - i} = \underset{i = 0}{\sum^{n - 1}} a_{i} {(b + c)}^{n - 1 - i}$ $\underset{i = 0}{\sum^{n - 1}} {a_{i} c^{n - 1 - i} + a_{i} b^{n - 1 - i} - a_{i} {(b + c)}^{n - 1 - i}} = 0$ $\underset{i = 0}{\sum^{n - 1}} [a_{i} {c^{n - 1 - i} + b^{n - 1 - i} - {(b + c)}^{n - 1 - i}}] = 0$ $\underset{i = 0}{\sum^{n - 3}} [a_{i} {c^{n - 1 - i} + b^{n - 1 - i} - {(b + c)}^{n - 1 - i}}] + a_{n - 2} {c^{n - 1 - n + 2} + b^{n - 1 - n + 2} - {(b + c)}^{n - 1 - n + 2}} + a_{n - 1} {c^{n - 1 - n + 1} + b^{n - 1 - n + 1} - {(b + c)}^{n - 1 - n + 1}} = 0$ $\underset{i = 0}{\sum^{n - 3}} [a_{i} {c^{n - 1 - i} + b^{n - 1 - i} - {(b + c)}^{n - 1 - i}}] + a_{n - 2} {c + b - (b + c)} + a_{n - 1} {1 + 1 - 1} = 0$ $\underset{i = 0}{\sum^{n - 3}} [a_{i} {c^{n - 1 - i} + b^{n - 1 - i} - {(b + c)}^{n - 1 - i}}] + a_{n - 1} = 0$ $a_{n - 1} = \underset{i = 0}{\sum^{n - 3}} [a_{i} {{(b + c)}^{n - 1 - i} - c^{n - 1 - i} - b^{n - 1 - i}}]$ $A c c o r d i n g t o t h e B i n o m i a l T h e o r e m,$ ${(b + c)}^{n - 1 - i} = \underset{k = 0}{\sum^{n - 1 - i}} (\begin{matrix} n - 1 - i \\ k \end{matrix}) b^{n - 1 - i - k} c^{k}$ $o r {(b + c)}^{n - 1 - i} = b^{n - 1 - i} + c^{n - 1 - i} + \underset{k = 1}{\sum^{n - 2 - i}} (\begin{matrix} n - 1 - i \\ k \end{matrix}) b^{n - 1 - i - k} c^{k} .$ $∴ {(b + c)}^{n - 1 - i} - b^{n - 1 - i} - c^{n - 1 - i} = \underset{k = 1}{\sum^{n - 2 - i}} (\begin{matrix} n - 1 - i \\ k \end{matrix}) b^{n - 1 - i - k} c^{k} .$ $a_{n - 1} = \underset{i = 0}{\sum^{n - 3}} [a_{i} {\underset{k = 1}{\sum^{n - 2 - i}} (\begin{matrix} n - 1 - i \\ k \end{matrix}) b^{n - 1 - i - k} c^{k}}]$ $a_{n - 1} = a_{0} {\underset{k = 1}{\sum^{n - 2}} (\begin{matrix} n - 1 \\ k \end{matrix}) b^{n - 1 - k} c^{k}} + a_{1} {\underset{k = 1}{\sum^{n - 3}} (\begin{matrix} n - 2 \\ k \end{matrix}) b^{n - 2 - k} c^{k}}$ $+ a_{2} {\underset{k = 1}{\sum^{n - 4}} (\begin{matrix} n - 3 \\ k \end{matrix}) b^{n - 3 - k} c^{k}} + . . . + a_{n - 4} {\underset{k = 1}{\sum^{2}} (\begin{matrix} 3 \\ k \end{matrix}) b^{3 - k} c^{k}} + a_{n - 3} {\underset{k = 1}{\sum^{1}} (\begin{matrix} 2 \\ k \end{matrix}) b^{2 - k} c^{k}}$ $O b s e r v e t h a t t h e f o r m o f t h e a b o v e e q u a t i o n$ $i s i n d e p e d e n t o f a_{n - 2}; s o a_{n - 2} i s a f r e e$ $v a r i a b l e s u c h t h a t 0 ⩽ a_{n - 2} ⩽ m i n (c - 1, b - 1) .$ $- - - - - - - - - - - - - - - - - - - - - - - - - - - - -$ $P r o p o s i t i o n : F o r a n y n ⩾ 3, b c d i v i d e s \underset{k = 1}{\sum^{n - 2}} (\begin{matrix} n - 1 \\ k \end{matrix}) b^{n - 1 - k} c^{k} .$ $Proof (induction) : B a s e c a s e n = 3 : \underset{k = 1}{\sum^{3 - 2}} (\begin{matrix} 3 - 1 \\ k \end{matrix}) b^{3 - 1 - k} c^{k} = (\begin{matrix} 2 \\ 1 \end{matrix}) b c$ $\Rightarrow b c d i v i d e s \underset{k = 1}{\sum^{n - 2}} (\begin{matrix} n - 1 \\ k \end{matrix}) b^{n - 1 - k} c^{k} f o r n = 3 .$ $I n d u c t i v e s t e p : S u p p o s e b c d i v i d e s \underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k \end{matrix}) b^{j - 1 - k} c^{k} f o r n = j .$ $i . e \underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k \end{matrix}) b^{j - 1 - k} c^{k} = r b c f o r s o m e r \in N .$ $F o r n = j + 1, w e h a v e t h a t \underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = \underset{k = 1}{\sum^{j - 1}} {(\begin{matrix} j - 1 \\ k \end{matrix}) + (\begin{matrix} j - 1 \\ k - 1 \end{matrix})} b^{j - k} c^{k}$
Commented by Yozzia last updated on 20/Aug/16

$L e t t (n) = \underset{k = 1}{\sum^{n - 2}} (\begin{matrix} n - 1 \\ k \end{matrix}) b^{n - 1 - k} c^{k}, n ⩾ 3 .$ $S i n c e b c d i v i d e s t (n) f o r n ⩾ 3, t h e n$ $b c d i v i d e s a_{n - 1} s i n c e a_{n - 1} = a_{0} t (n) + a_{1} t (n - 1) + a_{2} t (n - 2) + a_{3} t (n - 3) + . . . + a_{n - 3} t (3) .$ $N o w, b > 1, c > 1 a n d a l l a_{i} (i = 0, 1, 2, . . .)$ $a r e n o n - n e g a t i v e i n t e g e r s, w h e r e m i n (a_{0}) = 1 .$ $∴ a_{n - 1} = {b c l a}_{0} + . . . (l \in N) . B u t, i f m i n (a_{0}) = 1 \Rightarrow a_{n - 1} > b a n d a_{n - 1} > c,$ $w h i l e 0 ⩽ a_{n - 1} ⩽ m i n (c - 1, b - 1) < m i n (c, b) .$ $T h i s c o n t r a d i c t i o n i n d i c a t e s t h a t q$ $c a n n o t b e a n u m b e r h a v i n g d i g i t l e n g t h$ $n ⩾ 3, a n d q_{c} + q_{b} = q_{b + c} . / / / /$
Commented by Yozzia last updated on 20/Aug/16
$Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =b{Σ_(k=1) ^(j−2) (((j−1)),(k) ) b^(j−1−k) c^k + (((j−1)),((j−1)) ) b^0 c}+Σ_(k=1) ^(j−1) (((j−1)),((k−1)) ) b^(j−k) c^k Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =b×rbc+bc+Σ_(k=1) ^(j−1) (((j−1)),((k−1)) ) b^(j−k) c^k Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =bc(br+1)+{Σ_(k=1) ^(j−2) (((j−1)),((k−1)) ) b^(j−k) c^k + (((j−1)),((j−2)) ) b^(j−j+1) c^(j−1) } Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +{Σ_(k=1) ^(j−2) (((j−1)),((k−1)) ) b^(j−k) c^k } Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +b^(j−1) c+{c (((j−1)),(1) ) b^(j−2) c+c (((j−1)),(2) ) b^(j−3) c^2 +...+c (((j−1)),((j−3)) ) b^2 c^(j−3) + (((j−1)),((j−2)) ) bc^(j−2) − (((j−1)),((j−2)) ) bc^(j−2) } Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +b^(j−1) c+c{ (((j−1)),(1) ) b^(j−2) c+ (((j−1)),(2) ) b^(j−3) c^2 +...+ (((j−1)),((j−3)) ) b^2 c^(j−3) + (((j−1)),((j−2)) ) bc^(j−2) }− (((j−1)),((j−2)) ) bc^(j−1) Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +b^(j−1) c+cΣ_(k=1) ^(j−2) (((j−1)),(k) ) b^(j−1−k) c^k −(j−1) bc^(j−1) Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k =bc(br+1+cr)+b^(j−1) c Since j≥3⇒ j−1>1 ⇒bc divides b^(j−1) c. Therefore, bc divides Σ_(k=1) ^(j−1) ((j),(k) ) b^(j−k) c^k , if bc divides Σ_(k=1) ^(j−2) (((j−1)),(k) ) b^(j−1−k) c^k . Hence, by P.M.I, the proposition is true. −−−−−−−−−−−−−−−−−−−−−−−−−−−$
$\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b {\underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k \end{matrix}) b^{j - 1 - k} c^{k} + (\begin{matrix} j - 1 \\ j - 1 \end{matrix}) b^{0} c} + \underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j - 1 \\ k - 1 \end{matrix}) b^{j - k} c^{k}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b \times r b c + b c + \underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j - 1 \\ k - 1 \end{matrix}) b^{j - k} c^{k}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b c (b r + 1) + {\underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k - 1 \end{matrix}) b^{j - k} c^{k} + (\begin{matrix} j - 1 \\ j - 2 \end{matrix}) b^{j - j + 1} c^{j - 1}}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b c (b r + 1) + (j - 1) {b c}^{j - 1} + {\underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k - 1 \end{matrix}) b^{j - k} c^{k}}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b c (b r + 1) + (j - 1) {b c}^{j - 1} + b^{j - 1} c + {c (\begin{matrix} j - 1 \\ 1 \end{matrix}) b^{j - 2} c + c (\begin{matrix} j - 1 \\ 2 \end{matrix}) b^{j - 3} c^{2} + . . . + c (\begin{matrix} j - 1 \\ j - 3 \end{matrix}) b^{2} c^{j - 3} + (\begin{matrix} j - 1 \\ j - 2 \end{matrix}) {b c}^{j - 2} - (\begin{matrix} j - 1 \\ j - 2 \end{matrix}) {b c}^{j - 2}}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b c (b r + 1) + (j - 1) {b c}^{j - 1} + b^{j - 1} c + c {(\begin{matrix} j - 1 \\ 1 \end{matrix}) b^{j - 2} c + (\begin{matrix} j - 1 \\ 2 \end{matrix}) b^{j - 3} c^{2} + . . . + (\begin{matrix} j - 1 \\ j - 3 \end{matrix}) b^{2} c^{j - 3} + (\begin{matrix} j - 1 \\ j - 2 \end{matrix}) {b c}^{j - 2}} - (\begin{matrix} j - 1 \\ j - 2 \end{matrix}) {b c}^{j - 1}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b c (b r + 1) + (j - 1) {b c}^{j - 1} + b^{j - 1} c + c \underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k \end{matrix}) b^{j - 1 - k} c^{k} - (j - 1) {b c}^{j - 1}$ $\underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k} = b c (b r + 1 + c r) + b^{j - 1} c$ $S i n c e j ⩾ 3 \Rightarrow j - 1 > 1 \Rightarrow b c d i v i d e s b^{j - 1} c .$ $T h e r e f o r e, b c d i v i d e s \underset{k = 1}{\sum^{j - 1}} (\begin{matrix} j \\ k \end{matrix}) b^{j - k} c^{k}, i f b c d i v i d e s \underset{k = 1}{\sum^{j - 2}} (\begin{matrix} j - 1 \\ k \end{matrix}) b^{j - 1 - k} c^{k} .$ $H e n c e, b y P . M . I, t h e p r o p o s i t i o n i s t r u e .$ $- - - - - - - - - - - - - - - - - - - - - - - - - - -$
Commented by Rasheed Soomro last updated on 20/Aug/16

$W o w! W h a t a d e p t h y o u h a v e i n m a t h e m a t i c s!$ $G o o d w i s h e s f o r y o u a n d y o u r f u t u r e!$
Commented by Yozzia last updated on 20/Aug/16

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