Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)7 .
Is that a power?
(a+b+c)^7
?
Haven't worked through all the details, but I think if you consider that the sum of powers of all the terms in the expansion is 7, just examine every combination.
Naturally, all the terms with a^x b^y c^z are divisible by abc.
Consider a^7
a^2 = mc
c^2 = nb
a^7 = a*(a^2)*a^2
= a*(mc)^2*a^2
= a*m^2c^2*a^2
= a*m^2*nb*a^2
= abc*m^2*n*a^2
so, it appears that all of the terms will be divisible by abc.
Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7
To prove that abc divides (a + b + c)^7, we need to show that the remainder when (a + b + c)^7 is divided by abc is zero.
First, we can rewrite (a + b + c)^7 using the binomial theorem:
(a + b + c)^7 = C(7,0)a^7 + C(7,1)a^6b + C(7,2)a^5b^2 + C(7,3)a^4b^3 + C(7,4)a^3b^4 + C(7,5)a^2b^5 + C(7,6)ab^6 + C(7,7)b^7
where C(n,k) denotes the binomial coefficient "n choose k."
Now, we can see that each term in the expanded form of (a + b + c)^7 has either a, b, or c raised to a power equal or greater than 1. This is because if any of the variables (a, b, or c) were raised to a power of 0 or less in a term, it would not contribute to the remainder when divided by abc.
Let's focus on each term individually:
1. C(7,0)a^7: Since a divides a^7 (a is a factor of itself), this term is divisible by a.
2. C(7,1)a^6b: Since a divides a^6 and b divides b (b is a factor of itself), this term is divisible by ab.
3. C(7,2)a^5b^2: Since a divides a^5 and b^2 divides b^2, this term is divisible by a * b^2.
4. C(7,3)a^4b^3: Since a^4 divides a^4 and b^3 divides b^3, this term is divisible by a^4 * b^3.
5. C(7,4)a^3b^4: Similarly, this term is divisible by a^3 * b^4.
6. C(7,5)a^2b^5: This term is divisible by a^2 * b^5.
7. C(7,6)ab^6: This term is divisible by a * b^6.
8. C(7,7)b^7: Since b divides b^7, this term is divisible by b.
From this analysis, we can see that each term of the expanded form of (a + b + c)^7 is divisible by abc.
Therefore, we can conclude that abc divides (a + b + c)^7.