Find $a,b,c,d,e$ such that $\frac{s}a+1,\frac{s}b+1,\frac{s}c+1,\frac{s}d+1,\frac{s}e+1$ are all perfect squares $ (s=abcde)$

Are there five distinct positive integers $a,b,c,d,e$ such that $\dfrac{s}a+1,\dfrac{s}b+1,\dfrac{s}c+1,\dfrac{s}d+1,\dfrac{s}e+1$ are all perfect squares ? $ (s=abcde)$

If $a=1,b=2,c=12,d=2380,s=abcd=57120,$ then $\dfrac{s}a+1,\dfrac{s}b+1,\dfrac{s}c+1,\dfrac{s}d+1$ are all perfect squares. But the case of five numbers is more difficult than four numbers.

Solution 1:

Small contribution: if you drop the requirement that $e$ must be distinct from $a,b,c,d$ (but leave the rest the same), then the only tuples that I could find by brute force with $1+abcd\leq (2\times10^4)^2$ and some fairly large limit on $e$ are $$ \begin{array}{ccc} \{2,6,12,35,2\}, &\{1,2,12,1190,2\}, &\{2,4,68,105,2\}, \\\{2,10,87,156,2\}, &\{2,22,86,195,2\}, &\{2,23,90,204,2\}, \\\{2,6,247,330,2\}, &\{2,30,99,238,2\}, &\{2,58,129,360,2\}, \\\{2,52,161,396,2\}, &\{2,10,714,893,2\}, &\{2,35,1518,2014,2\}, \\\{2,12,3185,3588,2\}. \end{array} $$ Naturally, this doesn't answer your question.

Solution 2:

Denote

$bcde+1=A^2$, $\quad$ $acde+1=B^2$, $\quad$ $abde+1=C^2$, $\quad$ $abce+1=D^2$, $\quad$ $abcd+1=E^2$.

If $a<b<c<d<e$, then $A>B>C>D>E$.

My search was unsuccessful. I found only few "semi-solutions" (where $A,B,C$ are integer, but $D$ or $E$ isn't): here they are:

\begin{array}{|rrrrr|rrrrr|} \hline a & b & c & d & e & A & B & C & D & E \\ \hline 1 & 3 & 6 & 14 & 20 & 71 & 41 & 29 & 19 & \color{#CC3300}{(15.90...)} \\ 1 & 2 & 12 & 34 & 70 & 239 & 169 & 69 & 41 & \color{#CC3300}{(28.58...)} \\ 1 & 2 & 12 & 35 & 68 & 239 & 169 & 69 & \color{#CC3300}{(40.41...)} & 29 \\ 2 & 6 & 11 & 14 & 76 & 265 & 153 & 113 & \color{#CC3300}{(100.16...)} & 43 \\ 2 & 4 & 87 & 308 & 615 & 8119 & 5741 & 1231 & \color{#CC3300}{(654.24...)} & 463 \\ 11 & 34 & 39 & 140 & 464 & 9281 & 5279 & 4929 & \color{#CC3300}{(2601.51...)} & 1429 \\ 23 & 82 & 360 & 462 & 494 & 82081 & 43471 & 20747 & \color{#CC3300}{(18314.09...)} & 17711 \\ 11 & 28 & 195 & 792 & 4978 & 146719 & 91961 & 34847 & 17291 & \color{#CC3300}{(6896.92...)} \\ 4 & 182 & 395 & 957 & 2968 & 451879 & 66991 & 45473 & \color{#CC3300}{(29214.34...)} & 16589 \\ 7 & 470 & 912 & 2669 & 154548 & 13296959 & 1622753 & 1164941 & \color{#CC3300}{(680968.56...)} & 89489 \\ 2 & 385 & 403 & 1836 & 23240614 & 81366011 & 5864455 & 5731991 & \color{#CC3300}{(2685478.52...)} & 23869 \end{array}

And I have a doubt if solutions exist (at all) ...

Skeleton of search algorithm can be as follow:

Denote $M = cde$.

First, we need to find such triples $(a,b,M)$, that:
$a\cdot M +1 = B^2$,
$b\cdot M +1 = A^2$.

It is more convenient to talk about triples $(a,b,A)$ (where $A$ is the greatest square base).

To avoid brute-forcing, we observe that for many pairs $(a,b)$ there exist sequence $\{A_n \}$ (and, of course, $\{B_n\}, \{M_n\}$). And it is not hard to build/construct such sequence (for some reasonable limit), because exists a recurrence.

For example, consider $(a,b)=(1,10)$:

\begin{array}{|c|rrrrrrrrr} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline A_n & 9 & 41 & 79 & 351 & 1559 & 3001 & 13329 & 59201 & 113959 \\ \hline B_n & 3 & 13 & 25 & 111 & 493 & 949 & 4215 & 18721 & 36037 \\ \hline M_n & 8 & 168 & 624 & 12320 & 243048 & 900600 & 17766224 & 350475840 & 1298665368 \\ \hline \end{array}

We observe, that (here step is $3$): $$ A_{n+3} = 38 \cdot A_{n} - A_{n-3}, \qquad \qquad (n>3).\tag{1} $$ Yes, $3001 = 38 \cdot 79 - 1, 13329 = 38 \cdot 351 - 9, \ldots.$

Each pair $(a,b)$ has its own step, and its own multiplier near $A_n$.

A little bit more: we can define $A_0 = - A_{1}, A_{-1}=-A_2, A_{-2} = -A_3, \ldots$. Then formula $(1)$ will be true for any $n\in \mathbb{Z}$.

So, method is:

• considering different pairs $(a,b)$, find first terms of sequence $\{A_n \}$
• construct other terms (using recurrence, not bruting)
• for each triple $(a,b,A_n)$ consider all divisors of $A_n (c<d<e)$, and check them ...