Is that a power?
(a+b+c)^7
?
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 .
3 answers
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.
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