Problem of the Day B

Number Theory Level 4

$\large 2^{a} + 2^{b} = c!$

Given that $a, b, c$ are non-negative integers, how many ordered pairs $( a, b)$ are there such that the equation above is satisfied?

This problem is not original.

The answer is 5.

2 solutions

Leonel Castillo
Jul 2, 2018

I will prove that a solution is impossible for $c \geq 5$ . If such a solution existed then $2^a + 2^b \equiv 0 \mod 3 \implies (-1)^a + (-1)^b \equiv 0 \mod 3$ . 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 $2^a + 2^b \equiv 0 \mod 5$ . Because we know the parity of $a,b$ let's compute the even and odd powers of $2$ modulus 5:

$2^0 \equiv 1 \mod 5 \\ 2^2 \equiv 4 \mod 5$ $2^1 \equiv 2 \mod 5 \\ 2^3 \equiv 3 \mod 5$

Notice that opposites belong to the same parity. So if $a,b$ have opposite parity, there are no solutions to the equation $2^a + 2^b \equiv 0 \mod 5$ , a contradiction. We can now assume $c \leq 4$ , turning the problem into a routine computation.

Exactly I approached this way.

Taisanul Haque - 2 years, 7 months ago

Niranjan Khanderia
Feb 7, 2015

$\Large[1] ~~~~~~~~~ 2^0+2^0= 2!, \\ \Large [2-3]~~~ 2^1+2^2 =3!=2^2+2^1,\\ \Large [4-5] ~~~2^3+2^4=4!=2^4+2^3.$

How would you prove there exist no more?

Debajyoti Nandy - 6 years, 4 months ago

Let $d=b-a \geq 0$ We have $c!=2^a+2^b =2^a(1+2^{b-a})=2^a(1+2^d)$ .

If $c \geq 5$ , then $5\mid c! = 2^a(1+2^d)$ 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 $4n+2$ for some natural number $n$ .

If $c \geq 5$ , then $3\mid c! =2^a(1+2^d)$ 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 $2k+1$ for some natural number $k$ .

$d$ cannot be both $4n+2$ and $2k+1$ due to the following.

$d=d$

$2k+1=4n+2$

$2k-4n=1$

$2(k-2n)=1$

$k-2n=\frac{1}{2}$

$k-2n$ cannot equal $\frac{1}{2}$ for natural numbers $n$ and $k$ .

Since a contradiction exists that stops $5\mid c!$ and $3\mid c!$ then no solutions with $c \geq 5$ exist.

Jason Hughes - 6 years, 3 months ago

Will you please explain how for 5 d is even , but odd for 3 ?Thanks.

Niranjan Khanderia - 6 years, 3 months ago

For c>4, c! will have 0 at unit place. If you form a table of c! say up to c=10 and $2^{22}$ , 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.

Niranjan Khanderia - 6 years, 4 months ago

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

Trishita Barman - 11 months, 2 weeks ago

Problem of the Day B

This problem is not original.

The answer is 5.

2 solutions

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