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Source: Azerbaijan third round 2020

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Functional_equation

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Functional_equation

#1 h Jun 6, 2020, 9:16 AM • 1 Y

Y by Mango247

$a,b,c$ are positive integer.
Solve the equation:
$2^{a!}+2^{b!}=c^3$

This post has been edited 1 time. Last edited by Functional_equation, Jun 9, 2020, 6:15 AM

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1407 posts

#2 h Jun 6, 2020, 9:48 AM • 5 Y

Y by A-Thought-Of-God, ehuseyinyigit, Mango247, Mango247, Mango247

If $a,b\ge 3, 6|a!,b!\Rightarrow 2^{a!}\equiv 2^{b!}\equiv1\pmod 9$ so $2^{a!}+2^{b!}\equiv 2\pmod 9$ which is absurd.WLOG $a=2,1$ , $\pmod9$ again implies $b<3$ looking at all the cases we get the only solution as $a=b=2$ .

This post has been edited 3 times. Last edited by leibnitz, Jun 6, 2020, 9:50 AM

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Functional_equation

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Functional_equation

#3 h Jun 6, 2020, 9:59 AM

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My Solution

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#6 h Jun 25, 2021, 1:04 PM • 1 Y

Y by guywholovesmathandphysics

Case 1 = If a=b then 2^b!+1=c^3 and c must be in form of 2^a. From here we get 3a=b! +1 . So b<3 . By trying 1 and 2. For case 1 a=b=2 is the only solution.
Case 2. Becuse a and b symetric in this equation , we can take a>b .
Now we get 2^b!(2^(a!-b!) +1)=c^3 Because the part in the paranthesis doesn't divide 2, b! is divisible by 3.That makes b=3 or b>3. So a>3. Now outside of the paranthesis is cubic. So let's say 2^(a!-b!)+1= k^3 where k is positive int.
In modulo 7 x^3 can't be 2. But because a>3 and b is bigger or equal to 3. And a>b. a!-b! divides 3.
In modulo seven 2^(a!-b!) +1 is 2. We get a contradiction in this case.
So a and be must be equal and the only solution is (2,2,2) .

Azerbaycana Selamlar.

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lian_the_noob12

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lian_the_noob12

#7 h Mar 25, 2023, 6:52 PM

Y by

Mod 7 works too!!

This post has been edited 2 times. Last edited by lian_the_noob12, Mar 25, 2023, 6:55 PM

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#9 h Jul 31, 2023, 1:28 PM

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We have that $2^{a!} \equiv 2^{b!} \equiv 1\pmod{9}$ , if $a,b \ge 3$ , but this can't work because the sum is 2 mod 9 which is impossible. Now looking at the possible values, the only solution is $\boxed{a=b=c=2}.$

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F10tothepowerof34

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F10tothepowerof34

#10 h Aug 24, 2023, 9:21 PM

Y by

Clearly $2\mid c$ , thus $c^3\equiv -1\text{ or }1\pmod 7$ . Moreover notice that $2^{n!}\equiv 1\pmod 7, \forall n\ge3$
Therefore $2^{a!}+2^{b!}\equiv2\pmod 7, \forall a,b\ge3$ however this clearly contradicts the fact that $c^3\equiv-1\text{ or }1\pmod 7$ . Thus $a,b\le2$
Manually checking the cases yields $(a,b,c)=(2,2,2)$ to be the only viable solution $\blacksquare$ .

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#11 h Nov 17, 2023, 4:05 PM • 2 Y

Y by monoditetra, Leparolelontane

Also there is a question like that (it's from senior 2020);
pow(2,a!)+pow(2,b!)+pow(2,c!)=x³
First i investigated the possibilities;
i) a=b=c
3* pow(2,a!)=x³
x is consist of 2 and 3 divisors
We must to make pow(3,3k) but we can not do this. Because another divisor is pow(2,p). So there is no solution.
ii) The answer is asymmetric. So we will look one of them, and after finding all of the solutions we will write permutations of them. Assume that a=b<c
2*pow(2,a!)+pow(2,c!)=x³
pow(2,a!+1)+pow(2,c!)=x³ accept that :
c=a+1
pow(2,a!+1)*(1+2⁰)=pow(2,a!+2) And in the problem that's given that a€Z+ . It pays for a=1. And it has solution .
Click to reveal hidden text
$Permutations of (1,1,2)$
Let's look the equality like that;
a+1!=c and the equality will be:
1+ pow(2,c!-a!-1) it can not be pow(2,k) and also can not be a cube.
Let's look another case;
iii) a<b<c
pow(2,a!)*(1+pow(2,b!-a!)+pow(2,c!-a!)=x³
And also there are two cases. If it must be equal to a cube of the number, second factor must be pow(2,n). But it gives 1 to mod 2, so it can't.
Or a must be 3k and the second multiply must be cube.
Let's accept b!-a!=m and c!-a!=n
1+pow(2,m)+ pow(2,n)=y³
pow(2,m)+pow(2,n)=y³-1=(y-1)(y²+y+1)
y must be odd number. So y-1==>even number
And y²+y+1==>odd number. pow(2,m)+pow(2,n) there are only 2 factors. So it can not pay. It doesn't have solution.

Click to reveal hidden text
$Our answer: (1,1,2),(1,2,1),(2,1,1)$

This post has been edited 2 times. Last edited by zaidova, Feb 18, 2024, 7:59 PM

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#12 h Feb 18, 2024, 8:08 PM • 2 Y

Y by monoditetra, Leparolelontane

$a=b=c /is /the /only /solution /that/ we/ have.$

zaidova wrote:

Also there is a question like that (it's from senior 2020);
pow(2,a!)+pow(2,b!)+pow(2,c!)=x³
First i investigated the possibilities;
i) a=b=c
3* pow(2,a!)=x³
x is consist of 2 and 3 divisors
We must to make pow(3,3k) but we can not do this. Because another divisor is pow(2,p). So there is no solution.
ii) The answer is asymmetric. So we will look one of them, and after finding all of the solutions we will write permutations of them. Assume that a=b<c
2*pow(2,a!)+pow(2,c!)=x³
pow(2,a!+1)+pow(2,c!)=x³ accept that :
c=a+1
pow(2,a!+1)*(1+2⁰)=pow(2,a!+2) And in the problem that's given that a€Z+ . It pays for a=1. And it has solution .
Click to reveal hidden text
$Permutations of (1,1,2)$
Let's look the equality like that;
a+1!=c and the equality will be:
1+ pow(2,c!-a!-1) it can not be pow(2,k) and also can not be a cube.
Let's look another case;
iii) a<b<c
pow(2,a!)*(1+pow(2,b!-a!)+pow(2,c!-a!)=x³
And also there are two cases. If it must be equal to a cube of the number, second factor must be pow(2,n). But it gives 1 to mod 2, so it can't.
Or a must be 3k and the second multiply must be cube.
Let's accept b!-a!=m and c!-a!=n
1+pow(2,m)+ pow(2,n)=y³
pow(2,m)+pow(2,n)=y³-1=(y-1)(y²+y+1)
y must be odd number. So y-1==>even number
And y²+y+1==>odd number. pow(2,m)+pow(2,n) there are only 2 factors. So it can not pay. It doesn't have solution.

$Our answer: (1,1,2),(1,2,1),(2,1,1)$

Same idea if we search the cases like that, we will get a=b=c

$2^{a!}+2^{b!}=c^3$

> $2^{a!}+2^{a!}=a^3$
And that is equal to;

$2^{a!+1}=a^3$ $a=2=b=c$

This post has been edited 4 times. Last edited by zaidova, Nov 21, 2024, 2:00 PM

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#13 h Aug 31, 2024, 12:12 AM • 1 Y

Y by Atilla

WLOG $a\geq b$

1)If $b\geq 3$ , then $(2^{a!/3})^3+(2^{b!/3})^3=c^3$ , which is not true because of Fermat's Last Theorem
2)If $b=2$ , then:
a) $a\geq3\Rightarrow c^3-(2^{a!/3})^3=4$ No solution.
b) $a=2\Rightarrow 4+4=c^3$ (2,2,2)

3)If $b=1$ , then:
a) $a\geq3\Rightarrow c^3-(2^{a!/3})^3=2$ No solution.
b) $a=2\Rightarrow 4+2=c^3$ No solution.
c) $a=1\Rightarrow 2+2=c^3$ No solution.

This post has been edited 1 time. Last edited by Sadigly, Dec 18, 2024, 8:20 PM

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Atilla

#15 h Dec 13, 2024, 5:26 AM

Y by

Sadigly wrote:

WLOG $a\geq b$

1)If $b\geq 3$ , then $(2^{a!/3})^3+(2^{b!/3})^3=c^3$ , which is not true because of Fermat's Last Theorem
2)If $b=2$ , then:
a) $a\geq3\Rightarrow c^3-(2^{a!/3})^3=4$ No solution.
b) $a=2\Rightarrow 4+4=c^3$ (2,2,2)

3)If $b=1$ , then:
a) $a\geq3\Rightarrow c^3-(2^{a!/3})^3=1$ No solution.
b) $a=2\Rightarrow 4+2=c^3$ No solution.
c) $a=1\Rightarrow 2+2=c^3$ No solution.

Bro you made a mistake in b=1 a)
Here it's equals 2 not 1 but its still no solution

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5611 posts

#16 h Dec 13, 2024, 3:48 PM

Y by

The only solution is $\boxed{(2,2,2)}$ , which works. Now we show it's the only one.

Claim: If the sum of two powers of $2$ is a cube, then neither of the two powers of $2$ are $1 \pmod 7$ .
Proof: Suppose otherwise $2^x + 2^y = z^3$ for some nonnegative integers $x,y,z$ and that one of $2^x, 2^y$ are $1 \pmod 7$ . WLOG that $2^y \equiv 1 \pmod 7$ . Note that powers of $2$ are $1,2,$ or $4$ modulo $7$ , so $z^3$ is $2,3,$ or $5$ modulo $7$ , however cubes are either $-1,0,1$ modulo $7$ , absurd. $\square$

Thus, $6$ can't divide $a!$ or $b!$ , meaning $a,b$ are both less than $3$ . Checking cases ( $a=1,2$ and $b=1,2$ ) gives the desired result.

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#17 h Dec 18, 2024, 8:21 PM

Y by

Atilla wrote:

Bro you made a mistake in b=1 a)
Here it's equals 2 not 1 but its still no solution

Thanks,corrected

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1013 posts

#18 h Dec 29, 2024, 12:55 PM

Y by

This seems new...
Claim: $a = b$ .
Proof: Assume otherwise WLOG $a < b$ , and let $\frac{b!}{a!} = k$ .
Note that we then have
$2^{a!}(2^k+1) = c^3.$ In particular, $2^k+1$ is coprime to $2^{a!}$ , so it must itself be a perfect cube, say $m^3$ .
Then we have $2^k+1 = m^3 \implies m^2 + m + 1 \mid 2^k$ , but this isn't possible as $m^2 + m + 1$ is odd and not equal to $\pm 1$ . $\square$

Now we have $2^{a!+1} = c^3$ , which has a solution iff $a! \equiv 2 \pmod{3} \iff a = b = 2$ . This gives us the value $c = 2$ .
Thus the nly working solution is $\boxed{(a, b, c0 = (2, 2, 2)}$ .

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Aopsn

#19 h Jan 3, 2025, 12:45 PM

Y by

Fermat's little theorem kills this problem

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#20 h Mar 30, 2025, 5:34 PM

Y by

my solution:
i solved tis problem using modulo 3:
So if a,b<=2 we have that a=b=1:
so c^3=2+2=1 mod 3 <=> c=1 (fermat's litle theorem (a^p=a mod p))
Therefore a,b >=2. So a,b are even numbers.
It also means that 2^2k =1 mod 3
therefore a!=2s and b!=2t.
So 2^a!=1 mod 3 and 2^b! = 1 mod 3 <=> c^3 = 2 mod 3 <=> c=2 (fermat's litle theorem again)
Therefore we have that 2^a! 2^b!= 8 <=> a=b=2
So the solution is (a,b,c)=(2,2,2).
I Hope It Is Correct.

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