2 a + 2 b = c !
Given that a , b , c are non-negative integers, how many ordered pairs ( a , b ) are there such that the equation above is satisfied?
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Exactly I approached this way.
[ 1 ] 2 0 + 2 0 = 2 ! , [ 2 − 3 ] 2 1 + 2 2 = 3 ! = 2 2 + 2 1 , [ 4 − 5 ] 2 3 + 2 4 = 4 ! = 2 4 + 2 3 .
How would you prove there exist no more?
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Let d = b − a ≥ 0 We have c ! = 2 a + 2 b = 2 a ( 1 + 2 b − a ) = 2 a ( 1 + 2 d ) .
If c ≥ 5 , then 5 ∣ c ! = 2 a ( 1 + 2 d ) since c ! divides 5 and 2 a does not divide 5 then 1 + 2 d must divide 5 . For this to be true d must be of the form 4 n + 2 for some natural number n .
If c ≥ 5 , then 3 ∣ c ! = 2 a ( 1 + 2 d ) since c ! divides 3 and 2 a does not divide 3 then 1 + 2 d must divide 3 . For this to be true d must be of the form 2 k + 1 for some natural number k .
d cannot be both 4 n + 2 and 2 k + 1 due to the following.
d = d
2 k + 1 = 4 n + 2
2 k − 4 n = 1
2 ( k − 2 n ) = 1
k − 2 n = 2 1
k − 2 n cannot equal 2 1 for natural numbers n and k .
Since a contradiction exists that stops 5 ∣ c ! and 3 ∣ c ! then no solutions with c ≥ 5 exist.
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Will you please explain how for 5 d is even , but odd for 3 ?Thanks.
For c>4, c! will have 0 at unit place. If you form a table of c! say up to c=10 and 2 2 2 , there is no 2 a + 2 b that matches with c!. I only know some Theory of Numbers, some one knowing it can give a better proof.
2^1+2^2=6=3! 2^3+2^4=24=4! Exactly 4 pairs of (a,b) are there you can prove it by intuition too
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I will prove that a solution is impossible for c ≥ 5 . If such a solution existed then 2 a + 2 b ≡ 0 m o d 3 ⟹ ( − 1 ) a + ( − 1 ) b ≡ 0 m o d 3 . This implies that a has to be odd while b even (or vice-versa, but without loss of generality let's assume the mentioned configuration). Similarly, it must also be that 2 a + 2 b ≡ 0 m o d 5 . Because we know the parity of a , b let's compute the even and odd powers of 2 modulus 5:
2 0 ≡ 1 m o d 5 2 2 ≡ 4 m o d 5 2 1 ≡ 2 m o d 5 2 3 ≡ 3 m o d 5
Notice that opposites belong to the same parity. So if a , b have opposite parity, there are no solutions to the equation 2 a + 2 b ≡ 0 m o d 5 , a contradiction. We can now assume c ≤ 4 , turning the problem into a routine computation.