Finding consecutive numbers. If $p, q, r, s, t$ are consecutive numbers and  $p!+q!+r!+s^2=t^2$, then what are the values of $p, q, r, s$ and $ t$?
My try:
By brute force I got that $p=1$. But how to prove this more mathematically?
 A: Hint:
Note that  $t>s$  from the equation, so that  $q=p+1$,  $r=p+2$,  $s=p+3$, and  $t=p+4$. Rewriting the given equation as
$$p!(1+(p+1)+(p+1)(p+2))=t^2−s^2=t+s=2p+7,$$
we see that  $p∣7$. Hence  $p=1$  or  $p=7$. But  $p!>p(p−1)>2p−7$  for  $p=7$.
A: $$p!+q!+r!+s^2=t^2$$
$$p!(1+ q + qr) =(t - s)(s+t) = s + t$$
Since $s + t$ is odd,
both $p!$ and $(1+ q + qr)$ should be odd,
if $p!$ is odd then $p = 1$ because only odd factorial is $1!$
A: Without assuming that $p,q,r,s,t$ are in ascending order, we can still limit the possible solutions to effectively the one given. I'll use the form $p!+q!+r!=t^2-s^2$
First notice that one of the factorials on the LHS must be within $2$ of $t$, and $5!>7^2$ so certainly $t<7$ and the largest of $p,q,r$ is less than $5$.
So our options then are $A=\{2,3,4,5,6\}$, $B=\{1,2,3,4,5\}$, and if we allow zero, $C=\{0,1,2,3,4\}$.
For set $A$, $2!+3!+4! = 30 > 6^2-5^2$ will not work.
For set $B$,  $1!+2!+3!=9 = 5^2-4^2$ is good. Otherwise we would have $4!$ on the LHS and the difference of squares would not be feasible. So we get $\{p,q,r\}=\{1,2,3\}, s=4,t=5$
For set $C$,  $0!+1!+2!=4<4^2-3^2$; $0!+1!+3!=8<4^2-2^2$; $0!+2!+3!=9<4^2-1^2$ so there is no feasible solution there either.
