Find a, b, c, Such That a! b! = a! + b! + c!Date: 12/09/2003 at 06:15:46 From: Peter Subject: Is there a chance that a! b! = a! + b! + c! ? Find all triples of nonnegative integers a, b, c such that a! b! = a! + b! + c! I have no idea how to start. Would you please give me some hints? Date: 12/09/2003 at 12:07:24 From: Doctor Rob Subject: Re: Is there a chance that a! b! = a! + b! +c! ? Thanks for writing to Ask Dr. Math, Peter! This may not be the only way, or even the best way, but here is one way to attack this problem. Since the equation is symmetric in a and b, we can assume without loss of generality that a <= b. We can see that a = 0 and a = 1 are impossible, since b! = 1 + b! + c! is impossible. Also, if a = 2, then 2*b! = 2 + b! + c! b! - c! = 2 which is impossble. Thus b >= a >= 3. Then a! | b!, which implies that c! = 0 (mod a!), so a! | c!, so a <= c. Now divide the equation through by a!: b! = 1 + b!/a! + c!/a! 0 = 1 + 0 + c!/a! (mod b!/a!) c!/a! = -1 (mod b!/a!) That implies that either b!/a! = 1 or else b!/a! does not divide c!/a!, that is, either b = a or else b > c. Assume first that b > c. Then a!*b! - a! - b! = c! a!*b! - a! - b! + 1 = c! + 1 (a! - 1)*(b! - 1) = c! + 1 Since a >= 3, we must have 5 <= a! - 1, so 5*(b! - 1) <= (a! - 1)*(b! - 1) = c! + 1 <= (b-1)! + 1 5*b! <= (b-1)! + 6 <= 3*(b-1)! 5*b <= 3 which is again a contradiction. Thus b > c is impossible. That implies that a = b. Now we have a!^2 = 2*a! + c!, so a! | c! and a <= c. But a = c is impossible, since then one would have a!^2 = 3*a! a! = 3 a contradiction. Thus a < c, so a + 1 <= c. Divide through by a!: a! = 2 + c!/a! Now reduce modulo a + 1: a! = 2 (mod a + 1) If a + 1 is prime, then Wilson's Theorem says that a! = -1 (mod a + 1) -1 = 2 (mod a + 1) a + 1 | 3 a + 1 <= 3 another contradiction. Thus a + 1 must be composite. Now for almost all composite numbers, a + 1 | a!, which would imply that a + 1 | 2, another contradiction. Thus a + 1 must be a composite number with a + 1 not a divisor of a!. These are few. I leave the rest to you. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 12/10/2003 at 01:39:35 From: Peter Subject: Thank you Thank you very much! |
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