Easy and Nice

For discussing Olympiad Level Number Theory problems
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FahimFerdous
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Easy and Nice

Unread post by FahimFerdous » Thu Mar 07, 2013 12:14 am

Find all positive integers $m,n$ and primes $p$ where $p$ is greater than $3$, such that $m(4m^2+m+12)=3(p^n-1)$

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Tahmid Hasan
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Re: Easy and Nice

Unread post by Tahmid Hasan » Thu Mar 07, 2013 12:14 pm

$m(4m^2+m+12)=3(p^n-1) \Rightarrow (m^2+3)(4m+1)=3p^n$.
Note that $m=1$ is not a solution and $m^2+3>4m+1 \forall m>1$.
If $G.C.D.(m^2+3,4m+1)=1$, then $m^2+3=p^n,4m+1=3$ which doesn't provide any solution.
Now let $G.C.D.(m^2+3,4m+1)=d>1$.
$d|4(m^2+3)-m(4m+1)=12-m \Rightarrow d|4(12-m)+(4m+1)=49$.
So $d=7,49$. Clearly $p=7$.
CASE $1:G.C.D.(m^2+3,4m+1)=7$.
We have four sub-cases:$(m^2+3,4m+1)=(3.7,7^{n-1}),(7,3.7^{n-1}),(7^{n-1},3.7),(3.7^{n-1},7)$.
Solving these we get no solution.
CASE $2:G.C.D.(m^2+3,4m+1)=49$.
We again have four sub-cases:$(m^2+3,4m+1)=(3.7^2,7^{n-2}),(7^2,3.7^{n-2}),(7^{n-2},3,7^2),(3.7^{n-2},7^2)$.
Solving these we get one possible solution $(m,n)=(12,4)$ which indeed satisfies the given equation.
So $(m,n,p)=(12,4,7)$.
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