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### 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.

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

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)

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).

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/
```
Associated Topics:
High School Number Theory

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