Exponential Proof

Date: 03/06/2003 at 00:18:24
From: Abhay Dang
Subject: Algebra

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.

The binomial theorem doesn't help much. I tried putting

   b^2 = la
   c^2 = mb
   a^2 = nc

but I don't know what to do next.

Please help me. In your reply please give me a hint that may be 
helpful. Then I will try to solve this myself.

Thanks,
Abhay Dang

Date: 03/06/2003 at 04:13:04
From: Doctor Jacques
Subject: Re: Algebra

Hi Abhay,

First, note that the result is trivial if a=b=c=1, so we can exclude 
this case.

You should first prove that the prime factors of a, b, and c are the 
same, possibly with different exponents.

After that, it will be enough to prove that, for each such prime 
factor p, the exponent of p in abc is at most equal to the exponent 
of p in (a+b+c)^7.

Consider one of these prime factors, p.

Let the exponents of p in a, b, c, be k, m, n. We can write:

  k <= 2m
  m <= 2n
  n <= 2k

The exponent of p in (abc) is (k+m+n).

The exponent of p in (a+b+c)^7 is at least 7*min(k,m,n).

Can you continue from here? Please write back if you require further 
assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 03/07/2003 at 01:16:45
From: Abhay Dang
Subject: Algebra

Thanks, Dr. Jacques -

I have been able to transform your valuable hint into a proof. Please 
check whether the proof below needs changes. Also, I have a related 
doubt. Please clarify.

Proof

Let's consider any prime factor p of a. Since a divides b^2, therefore 
p must be a prime factor of b^2 as well (because otherwise b^2/a won't 
be an integer). Therefore p must be a prime factor of b as well. As b 
divides c^2 therefore p must be a prime factor of c.

This proves that all prime factors of a are prime factors of b and c.
Similarly, all prime factors of b are prime factors of a and c, and 
all prime factors of c are prime factors of b and a.

Hence all the prime factors of a,b,c are same. (Please check this 
statement. I think my English is not good enough.)

Now let's say the power of p in a,b,c is k,m,n, and k is the least of 
the three. As a divides b^2, therefore the power of p in a must not 
exceed the power of p in b^2.

Therefore

   2m >= k  (i)

Similarly 

   2n >= m  (ii)
   2k >= n  (iii)

Multiplying (iii) by 2

   4k >= 2n (iv)

Comparing (iv) with (ii)

   4k >= 2n >= m

Therefore

   4k >= m (v)

Adding (iii) and (v)

   6k >= m + n 

or

   7k >= k + m + n (I)

Now, the power of p in abc = power of p in a + power of p in b + 
power of p in c = k+m+n. Power of p in (a+b+c)^7 = 7 * Min(k,m,n) = 7k 
(by assumption).

For abc to divide (a+b+c)^7
power of p in abc <= power of p in (a+b+c)^7

i.e.

   k + m + n <= 7k (which we have proved in I)

Therefore abc divides (a + b + c)^7
Even if we assume m or n is least it won't make any difference because 
abc and (a+b+c)^7 are cyclic expressions.

End of Proof

Also I have one more doubt.
If in place of (a + b + c)^7 it had been (a + 2b + 3c)^7 or something 
else, will the proof hold true? Or if in place of abc it had been 2abc 
or abc/2. Should the proof hold true then?

Once again thanks for your extremely valuable help. No other math site 
equals Dr. Math.

Best regards,
Abhay Dang

Date: 03/07/2003 at 02:23:54
From: Doctor Jacques
Subject: Re: Algebra

Hi Abhay,

You proof is quite correct, congratulations. There is a minor 
inaccuracy, when you write:

>Power of p in (a+b+c)^7 = 7 * Min(k,m,n) = 7k (by assumption)

In fact, we have:

 Power of p in (a+b+c)^7 >= 7 * Min(k,m,n)

For example, if a=b=c=3, (a+b+c) is divisible by 3^2. This has no 
impact on the correctness of the proof (in some sense, it makes things 
even better).

Concerning your other questions:

If we have (a+2b+3c) instead of (a+b+c), the proof remains the same. 
Indeed if p^k divides a, b, and c, it will divide (a+2b+3c).

If we have 2abc instead of abc, the theorem is no longer true. The 
power of 2 in 2abc will be k+m+n+1 in this case. For example, if a, 
b, and c are odd, (a+b+c)^7 is also odd and cannot be divisible by 
the even number (2abc).

If we have (abc)/2 (assuming this is an integer) instead of abc, the 
the theorem still holds by transitivity:

  (abc)/2 | abc | (a+b+c)^7

Does this make sense?

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/