Problem 14:
If $m +12 = p^a$ and $m -12 = p^b$ where $a,b,m$ are integers and $p$ is a prime number. Find all possible primes $p > 0$ . [Note: $p$ only takes three values]
BdMO National Higher Secondary 2007/14
Re: BdMO National Higher Secondary 2007/14
This is our $500^{th}$ topic.
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- Tahmid Hasan
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Re: BdMO National Higher Secondary 2007/14
let's name the eqs $(i)$ and $(ii)$ and do $(i)-(ii)$
than divide three cases
1.$a>b$
2.$a<b$
3.$a=b$
than divide three cases
1.$a>b$
2.$a<b$
3.$a=b$
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- Cryptic.shohag
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Re: BdMO National Higher Secondary 2007/14
Actually it's not necessary as it's clear that $a>b$.Tahmid Hasan wrote:let's name the eqs $(i)$ and $(ii)$ and do $(i)-(ii)$
than divide three cases
1.$a>b$
2.$a<b$
3.$a=b$
$m+12=p^a$....(i)
$m-12=p^b$....(ii)
So, $p^a-p^b=24$. As $p^a$ and $p^b$ both are multiples of p, we can write $p^a-p^b=kp$ where k is an integer. So,
$kp=24$
As k is an integer, so p has to be a divisor of 24. Among the divisors of 24 we find 2 prime numbers and they are 2 and 3. So, p=2,3....
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein
- Cryptic.shohag
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Re: BdMO National Higher Secondary 2007/14
Oops!! I missed one solution. In last solution I considered only $a,b>0$....
As $a>b$ and $p^a-p^b=24$, a's value cannot be 0. So, if we consider $b=0$, then we get,
$p^a-1=24$
$p^a=25$
$p^a=5^2$....
So, another value of p is 5....
As $a>b$ and $p^a-p^b=24$, a's value cannot be 0. So, if we consider $b=0$, then we get,
$p^a-1=24$
$p^a=25$
$p^a=5^2$....
So, another value of p is 5....
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein
- Cryptic.shohag
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Re: BdMO National Higher Secondary 2007/14
So, p=2,3,5....
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein