A Problem from Dustan
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Find all $p,q,r,s>1$ positive integers such that $p!+q!+r!=2^s$.
Hmm..Hammer...Treat everything as nail
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- Posts:194
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Re: A Problem from Dustan
My SOLN(PLS CONFIRM):
Last edited by Asif Hossain on Mon May 03, 2021 10:35 pm, edited 2 times in total.
Hmm..Hammer...Treat everything as nail
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- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: A Problem from Dustan
o sorry, p=5 is also a possiblity for $\nu_{2}(p!)=3$
fixed now
Hmm..Hammer...Treat everything as nail
Re: A Problem from Dustan
Solution with some of my friends.
As well $r=2$ and there is no solution when $4\leq q$ and $p=q=r$
We will see two cases.
Case 1: $q=2$
The equation becomes
$p!+4=2^s$
Or,$0+1\cong (-1)^s$ (mod $3$)
If, $s$ is odd then it's a contradiction.
Let $s=2m$,$m>1$
So,
$p!+4=4^m$
If, $p>7$ then
L.H.S.$ \cong 4$ (mod $8$)
R.H.S.$\cong 0$(mod $8$)
Checking the values from $3 \, to \, 7$ there is no solution.
So, $q\neq 2$
Case 2: $q=3$
$p!+8=2^s$, $s>3$
Write $s=4+a$ [$a=0,1,2,....$]
$p!+8=16*2^a$
L.H.S.$\cong 8$( mod $16$), $p>5$
R.H..S.$\cong 0$ (mod $16$)
So the remain possibilities of $p$ are
$P=3,4,5$
The rest is trivial.
As well $r=2$ and there is no solution when $4\leq q$ and $p=q=r$
We will see two cases.
Case 1: $q=2$
The equation becomes
$p!+4=2^s$
Or,$0+1\cong (-1)^s$ (mod $3$)
If, $s$ is odd then it's a contradiction.
Let $s=2m$,$m>1$
So,
$p!+4=4^m$
If, $p>7$ then
L.H.S.$ \cong 4$ (mod $8$)
R.H.S.$\cong 0$(mod $8$)
Checking the values from $3 \, to \, 7$ there is no solution.
So, $q\neq 2$
Case 2: $q=3$
$p!+8=2^s$, $s>3$
Write $s=4+a$ [$a=0,1,2,....$]
$p!+8=16*2^a$
L.H.S.$\cong 8$( mod $16$), $p>5$
R.H..S.$\cong 0$ (mod $16$)
So the remain possibilities of $p$ are
$P=3,4,5$
The rest is trivial.