Finding consecutive numbers.

Question

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?

Answer 1

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 $p=7$.

Answer 2

$$p!+q!+r!+s^2=t^2$$

$$p!(1+ q + qr) =(t - s)(s+t) = s + t$$

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 $1!$

Answer 3

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.