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Factorial And Square

Problem

Given that a, b, and c are positive integers, solve the following equation.

a!b! = a! + b! + c2


Solution

Without loss of generality let us assume that b less than or equal a.

If b = a, then, (a!)2 = 2(a!) + c2.

therefore (a!)2 minus 2(a!) + 1 = c2 + 1
    (a! minus 1)2 = c2 + 1

However, as there exists no square which is one more than another square we deduce that b less than a.

In addition, b not equal 1, otherwise, a! = a! + 1 + c2 implies c2 = -1.

Dividing a!b! = a! + b! + c2 by b! we get, a! = a!/b! + 1 + c2/b!.

Clearly LHS is integer, and as a!/b! is integer, c2/b! must also be integer. Furthermore, RHS greater than or equal 3, which means that a! greater than or equal 3 implies a greater than or equal 3.

From a!b! = a! + b! + c2, we get a!b! minus a! minus b! + 1 = c2 + 1.

therefore (a! minus 1)(b! minus 1) = c2 + 1.

Let us assume that a prime, p congruent 3 mod 4 divides the RHS.

If c2 + 1 congruent 0 mod p, it follows that c2 congruent -1 mod p, and it is clear that p does not divide c.

therefore (c2)(pminus1)/2 = cpminus1 congruent (-1)(pminus1)/2 mod p.

But we are given that p congruent 3 mod 4, so pminus1 congruent 2 mod 4, which means that (pminus1)/2 will be odd, and (-1)(pminus1)/2 = -1.

Hence cpminus1 congruent -1 mod p, which is a contradiction, because HCF(c, p) = 1 and by Fermat's Little Theorem, cpminus1 congruent 1 mod p.

In other words, there exists no prime, p congruent 3 mod 4, which divides c2 + 1.

However, for a greater than or equal 4, a! congruent 0 mod 4, and a! minus 1 congruent 3 mod 4; that is, LHS is divisible by a prime, p congruent 3 mod 4.

Hence 1 less than b less than a less than or equal 3, but we have already established that a greater than or equal 3, so we deduce that a = 3 and b = 2.

From a!b! = a! + b! + c2, we get, 12 = 6 + 2 + c2 implies c = 2.

That is, the original equation has a unique solution: a = 3, b = c = 2.

Related problems:

Factorial Symmetry: a!b! = a! + b!

Factorial And Power Of 2: a!b! = a! + b! + 2c

Factorial Equation: a!b! = a! + b! + c!

Problem ID: 220 (30 Mar 2005)     Difficulty: 4 Star

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