Exponential ProofDate: 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/ |
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