A Problem from Dustan

For discussing Olympiad Level Number Theory problems
Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm
A Problem from Dustan

Unread post by Asif Hossain » Sun May 02, 2021 2:22 pm

Find all $p,q,r,s>1$ positive integers such that $p!+q!+r!=2^s$.
Hmm..Hammer...Treat everything as nail

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: A Problem from Dustan

Unread post by Asif Hossain » Sun May 02, 2021 2:54 pm

My SOLN(PLS CONFIRM):
We will do by casework.
Case 1: $p=q=r=x$
Not possible since $3 \perp 2$

Case 2: WLOG $p=q=x \neq r$
then $2x!+r!=2^s$ if $x>r$ then $r!(2k+1)=2^s$ which is clearly not possible.
So, $r>x$ then we would get $x!(2+\frac{r!}{x!})=2^s$
It is clear that $x=2$ either we would get other divisors than 2. So,it implies $r!=4(2^{s-2}-1)$
so,$\nu_{2}(r!)=2$ which is not possible.So, no solution either.

Case 3:WlOG $p>q>r$
then we can write $r!(\frac{p!}{r!}+\frac{q!}{r!}+1)=2^s$ again $r=2$ implying $p!+q!=2(2^{s-1}-1)$
From here we can get $\nu_{2}(q!)=1$ since $p>q$. So, either $q=2=r$ not possible or $q=3$ which yields $p!=8(2^{s-3}-1)$ so $\nu_{2}(p!)=3$ yielding the only possiblity $p=4,5$
So,the only solutions $(p,q,r)=(4,3,2),(5,3,2)$ and all their permutation and $s=5,7$.$\square$
Last edited by Asif Hossain on Mon May 03, 2021 10:35 pm, edited 2 times in total.
Hmm..Hammer...Treat everything as nail

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

Re: A Problem from Dustan

Unread post by Dustan » Sun May 02, 2021 2:59 pm

$p=5$ is also a soln

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: A Problem from Dustan

Unread post by Asif Hossain » Sun May 02, 2021 3:08 pm

Dustan wrote:
Sun May 02, 2021 2:59 pm
$p=5$ is also a soln
o sorry, p=5 is also a possiblity for $\nu_{2}(p!)=3$
fixed now :oops:
Hmm..Hammer...Treat everything as nail

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

Re: A Problem from Dustan

Unread post by Dustan » Sun May 02, 2021 6:15 pm

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.

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