Question

Let \[a,b,c\] be positive integers such that \[a\] divides \[{{b}^{2}}\], \[b\] divides \[{{c}^{2}}\] and \[c\] divides \[{{a}^{2}}\] then prove that \[abc\] divides \[{{\left( a+b+c \right)}^{7}}\]

Answer 1

Hint: We solve this problem by using the simple theorem of division. We use that if \[x\] divides \[y\] then the mathematical representation will be \[x|y\] then we use the given conditions to find the nature of divisors of \[a,b,c\] then, we use some standard conditions that are listed below
(1) If \[x|y\] and \[y|z\] then \[x|z\]
(2) If \[x|{{y}^{2}}\] then \[x|y\]
(3) If \[x|y\] then \[{{x}^{n}}|{{y}^{n}}\]
(4) If \[x|y,z|u\] then, \[u|y\] can also be written as \[x|z\]
(5) If \[{{x}^{n}}|{{y}^{k}}\] then, \[k\le n\]

Complete step by step answer:
We are given that \[a\] divides \[{{b}^{2}}\], \[b\] divides \[{{c}^{2}}\] and \[c\] divides \[{{a}^{2}}\]
Let us assume that there exists a prime number \['p'\] which divides \['a'\]
We know that if \[x\] divides \[y\] then the mathematical representation will be \[x|y\]
By using the above representation we can say that \[p|a\]
We know that if \[x|y\] and \[y|z\] then \[x|z\]
By using the above result we get that if \[p|a\] and \[a|{{b}^{2}}\] then \[p|{{b}^{2}}\]
Now, we know that if \[x|{{y}^{2}}\] then \[x|y\]
By using the above result we get that if \[p|{{b}^{2}}\] then \[p|b\]
We are given that \[b|{{c}^{2}}\] so, we can write that \[p|{{c}^{2}}\] and \[p|c\]
Similarly, we are given that \[c|{{a}^{2}}\] so, we can write that \[p|{{a}^{2}}\] and \[p|a\]
Here, we can see that the prime number \['p'\] divides all \[a,b,c\]
Similarly, we can say that there exists other prime divisor that divides all \[a,b,c\]
So, we can say that \[a,b,c\] has the same set of prime divisors.
Now, let us assume that there are three natural numbers \[q,r,s\in \mathbb{N}\] such that
\[{{p}^{q}}|a\] (That is \[{{p}^{q}}\] divides \[a\]) then we can say that \[{{p}^{2q}}|{{a}^{2}}\]
\[{{p}^{r}}|b\](That is \[{{p}^{r}}\] divides \[b\]) then we can say that \[{{p}^{2r}}|{{b}^{2}}\]
\[{{p}^{s}}|c\](That is \[{{p}^{s}}\] divides \[c\]) then we can say that \[{{p}^{2s}}|{{c}^{2}}\]
We are given that \[b|{{c}^{2}}\]
We know that if \[x|y,z|u\] then, \[u|y\] can also be written as \[x|z\]
By using the above statement we can represent \[b|{{c}^{2}}\] as \[{{p}^{2s}}|{{p}^{r}}\]
We know that if \[{{x}^{n}}|{{y}^{k}}\] then, \[k\le n\]
By using this statement to above equation we get
\[\Rightarrow r\le 2s.....equation(i)\]
We are given with other condition that is \[c|{{a}^{2}}\]
Similarly, by using the above conditions we used for \[b|{{c}^{2}}\] we get \[{{p}^{2q}}|{{p}^{s}}\] that is
\[\Rightarrow s\le 2q........equation(ii)\]
Now, by combining both equation (i) and equation (ii) we get
\[\begin{align}
  & \Rightarrow r\le 2s \\
 & \Rightarrow r\le 4q.......equation(iii) \\
\end{align}\]
Now, by adding the equation (ii) and equation (iii) we get
\[\begin{align}
  & \Rightarrow s+r\le 2q+4q \\
 & \Rightarrow q+r+s\le 7q \\
\end{align}\]
We assumed that
\[{{p}^{q}}|a\] (That is \[{{p}^{q}}\] divides \[a\])
\[{{p}^{r}}|b\](That is \[{{p}^{r}}\] divides \[b\])
\[{{p}^{s}}|c\](That is \[{{p}^{s}}\] divides \[c\])
Now, by combining all the three above statements we get the product \[abc\] as \[{{p}^{q+r+s}}|abc\]
Here, we have the maximum value of \[q+r+s\] as \[7q\] so, we can rewrite the above equation by using the condition that if \[{{x}^{n}}|{{y}^{k}}\] then, \[k\le n\] as \[{{p}^{7q}}|abc\]
Now, let us assume that the \['q'\] is the minimum value among \[q,r,s\in \mathbb{N}\] then from the conditions we assumed we can say that
\[\begin{align}
  & \Rightarrow {{p}^{q}}|\left( a+b+c \right) \\
 & \Rightarrow {{p}^{7q}}|{{\left( a+b+c \right)}^{7}} \\
\end{align}\]
Here we have the condition that \[q+r+s\le 7q\]

By using the condition that if \[{{x}^{n}}|{{y}^{k}}\] then, \[k\le n\] we get
\[\Rightarrow {{p}^{q+r+s}}|{{\left( a+b+c \right)}^{7}}\]
Here we got the condition that the number \['p'\] divides all \[a,b,c\]
So from the condition that is \[{{p}^{q+r+s}}|abc\] and \[{{p}^{q+r+s}}|{{\left( a+b+c \right)}^{7}}\] we get
\[\Rightarrow abc|{{\left( a+b+c \right)}^{7}}\]
Therefore, we can conclude that \[abc\] divides \[{{\left( a+b+c \right)}^{7}}\]
Hence the required result has been proved.

Note: We can explain the above question by taking the example.
We are given that \[a\] divides \[{{b}^{2}}\], \[b\]divides \[{{c}^{2}}\] and \[c\] divides \[{{a}^{2}}\]
One of the conditions that satisfies the given conditions is that
\[\Rightarrow a=b=c\]
Let us assume that
\[\Rightarrow a=b=c=k\]
Now the value of \[{{\left( a+b+c \right)}^{7}}\] can be written as
\[\Rightarrow {{\left( a+b+c \right)}^{7}}={{3}^{7}}\times {{k}^{7}}\]
Also we can write the value of \[abc\] as
\[\Rightarrow abc={{k}^{3}}\]
Here we can see that \[{{k}^{3}}|\left( {{3}^{7}}\times {{k}^{7}} \right)\]
By rewriting the above statement in terms of \[a,b,c\] we get
\[\Rightarrow abc|{{\left( a+b+c \right)}^{7}}\]
Therefore, we can conclude that \[abc\] divides \[{{\left( a+b+c \right)}^{7}}\]
Hence the required result has been proved.