## A^2 = B^3 + C^3

Solve, given all unknowns are non-zero integers

$A^2 \; = \; B^3 \; + \; C^3$
$B^2 \; = \; C^3 \; + \; D^3$
$C^2 \; = \; D^3 \; + \; E^3$
$D^2 \; = \; E^3 \; + \; F^3$
$E^2 \; = \; F^3 \; + \; G^3$
$F^2 \; = \; G^3 \; + \; A^3$

## Set {4,b,c,d,e} such that the product of any two of them increased by 1 is a square

The set   $\{ \, a, \; b, \; c, \; d, \; e \, \}$   has the property that the product of any two of them plus one is the square of a rational number.

$a \,b + 1$   ……….   $b \,c + 1$   ……….   $c \,d + 1$   ……….   $d \,e + 1$
$a \,c + 1$   ……….   $b \,d + 1$   ……….   $c \,e + 1$
$a \,d + 1$   ……….   $b \,e + 1$
$a \,e + 1$

are all perfect squares

Here we have   $a = 4$

## smallest integer whose first n multiples all contain a 3

To find the smallest integer whose first n multiples all contain a 3

$\{153, \; 306 \}$
153×1,   153×2

$\{1153, \; 2306, \; 3459 \}$
1153×1,   1153×2,   1153×3

$\{1183, \; 2366, \; 3549, \; 4732 \}$
1183×1,   1183×2,   1183×3,   1183×4

$\{3465, \; 6930, \; 10395, \; 13860, \; 17325 \}$
3465×1,   3465×2,   3465×3,   3465×4,   3465×5

$\{34566, \; 69132, \; 103698, \; 138264, \; 172830, \; 207396 \}$
34566×1,   34566×2,   34566×3,   34566×4,   34566×5,   34566×6

$\{53465, \; 106930, \; 160395, \; 213860, \; 267325, \; 320790, \; 374255 \}$
53465×1,   53465×2,   53465×3,   53465×4,   53465×5,   53465×6,   53465×7

$\{7673, \; 15346, \; 23019, \; 30692, \; 38365, \; 46038, \; 53711, \; 61384 \}$
7673×1,   7673×2,   7673×3,   7673×4,   7673×5,   7673×6,   7673×7,   7673×8

Can you find an integer whose first 9 multiples all contain a 3?

What about the smallest integer whose first 10 multiples all contain a 3?

——————————————————————

Paul’s solutions:

first 9 multiples of 65913
first 9 multiples of 76923

first 10 multiples of 65913
first 10 multiples of 76923

first 11 multiples of 65913
first 11 multiples of 76923

first 12 multiples of 76923

first 13 multiples of 232767
first 13 multiples of 257673

first 14 multiples of 232767
first 14 multiples of 257673

first 15 multiples of 232767
first 15 multiples of 257673

first 16 multiples of 232767

first 17 multiples of 232767

——————————————————————

first 18 multiples of 7692391

Can you find an integer whose first 19 multiples all contain a 3?
Can you find an integer whose first 20 multiples all contain a 3?

first 21 multiples of 2307767
first 22 multiples of 3076923
first 23 multiples of 6923313

Posted in Number Puzzles | Tagged | 4 Comments

## Set {3,b,c,d,e} such that the product of any two of them increased by 1 is a square

The set   $\{ \, a, \; b, \; c, \; d, \; e \, \}$   has the property that the product of any two of them plus one is the square of a rational number.

$a \,b + 1$   ……….   $b \,c + 1$   ……….   $c \,d + 1$   ……….   $d \,e + 1$
$a \,c + 1$   ……….   $b \,d + 1$   ……….   $c \,e + 1$
$a \,d + 1$   ……….   $b \,e + 1$
$a \,e + 1$

are all perfect squares

Here we have   $a = 3$

We obtain 2 families of solutions:

………………………………………………………………………………………

and

………………………………………………………………………………………

## Set {2,b,c,d,e} such that the product of any two of them increased by 1 is a square

The set   $\{ \, a, \; b, \; c, \; d, \; e \, \}$   has the property that the product of any two of them plus one is the square of a rational number.

$a \,b + 1$   ……….   $b \,c + 1$   ……….   $c \,d + 1$   ……….   $d \,e + 1$
$a \,c + 1$   ……….   $b \,d + 1$   ……….   $c \,e + 1$
$a \,d + 1$   ……….   $b \,e + 1$
$a \,e + 1$

are all perfect squares

where   $a = 2$

## Set {1,b,c,d,e} such that the product of any two of them increased by 1 is a square

The set   $\{ \, a, \; b, \; c, \; d, \; e \, \}$   has the property that the product of any two of them plus one is the square of a rational number.

where   $a = 1$

## Set {a,b,c,d,e} such that the product of any two of them increased by 1 is a square

The set   $\{ \, a, \; b, \; c, \; d, \; e \, \}$   has the property that the product of any two of them plus one is the square of a rational number.

where   $a = 1, 2, 3, 4, ...$

the next solutions are:

Find the next set.

## Set {a,b,c,d} such that the product of any two of them increased by 1 is a square — Part 3

$1 + a \,b = A^2$   ………………….   $1 + b \,c = D^2$
$1 + a \,c = B^2$   ………………….   $1 + b \,d = E^2$
$1 + a \,d = C^2$   ………………….   $1 + c \,d = F^2$

$a = 3$
$b = 3 \, n^2 - 4 \, n + 1$
$c = 3 \, n^2 + 2 \, n$
$d = 4 \, (27 \, n^4 - 18 \, n^3 - 12 \, n^2 + 5 \, n + 2)$

$1 + a \,b = (3 n - 2)^2$   ……………………   $1 + b \,c = (3 \, n^2 - n - 1)^2$
$1 + a \,c = (3 \, n + 1)^2$   ……………………   $1 + b \,d = (18 \, n^3 - 18 \, n^2 - 2 \, n + 3)^2$
$1 + a \,d = (18 \, n^2 - 6 \, n - 5)^2$   ………..   $1 + c \,d = (18 \, n^3 - 8 \, n - 1)^2$

Or

$a = 3$
$b = 3 \, n^2 - 2 \, n$
$c = 3 \, n^2 + 4 \, n + 1$
$d = 4 \, (27 \, n^4 + 18 \, n^3 - 12 \, n^2 - 5 \, n + 2)$

$1 + a \,b = (3 \, n - 1)^2$   ……………………   $1 + b \,c = (3 \, n^2 + n - 1)^2$
$1 + a \,c = (3 \, n + 2)^2$   ……………………   $1 + b \,d = (18 \, n^3 - 8 \, n + 1)^2$
$1 + a \,d = (18 \, n^2 + 6 \, n - 5)^2$   ………..   $1 + c \,d = (18 \, n^3 + 18 \, n^2 - 2 \, n - 3)^2$

…………………………………………………………..

$a = 4$

$a = 4$
$b = n^2 + n$
$c = n^2 + 5 \, n + 6$
$d = 4 \, (4 \, n^4 + 24 \, n^3 + 45 \, n^2 + 27 \, n + 5)$

$1 + a \,b = (2 n + 1)^2$   ……………………………..   $1 + b \,c = (n^2 + 3 \, n + 1)^2$
$1 + a \,c = (2 n + 5)^2$   ……………………………..   $1 + b \,d = (4 \, n^3 + 14 \, n^2 + 10 \, n + 1)^2$
$1 + a \,d = (8 n^2 + 24 n + 9)^2$   ………………….   $1 + c \,d = (4 \, n^3 + 22 \, n^2 + 34 \, n + 11)^2$

…………………………………………………………..

$a = 5$

$a = 5$
$b = 5 \, n^2 - 8 \, n + 3$
$c = 5 \, n^2 + 2 \, n$
$d = 4 \, (125 \, n^4 - 150 \, n^3 + 27 \, n + 4)$

$1 + a \,b = (5 \, n - 4)^2$   ……………………….   $1 + b \,c = (5 \, n^2 - 3 \, n - 1)^2$
$1 + a \,c = (5 \, n + 1)^2$   ……………………….   $1 + b \,d = (50 \, n^3 - 70 \, n^2 + 14 \, n + 7)^2$
$1 + a \,d = (50 \, n^2 - 30 \, n - 9)^2$   ………….   $1 + c \,d = (50 \, n^3 - 20 \, n^2 - 16 \, n - 1)^2$

$10 \, (1 + b \,c) \; + \; 1 \; = \; 1 \; + \; a \,d$

Or

$a = 5$
$b = 5 \, n^2 - 2 \, n$
$c = 5 \, n^2 + 8 \, n + 3$
$d = 4 \, (125 \, n^4 + 150 \, n^3 - 27 \, n + 4)$

$1 + a \,b = (5 \, n - 1)^2$   ……………………….   $1 + b \,c = (5 \, n^2 + 3 \, n - 1)^2$
$1 + a \,c = (5 \, n + 4)^2$   ……………………….   $1 + b \,d = (50 \, n^3 + 20 \, n^2 - 16 \, n + 1)^2$
$1 + a \,d = (50 \, n^2 + 30 \, n - 9)^2$   ………….   $1 + c \,d = (50 \, n^3 + 70 \, n^2 + 14 \, n - 7)^2$

$10 \, (1 + b \,c) \; + \; 1 \; = \; 1 \; + \; a \,d$

## Set {a,b,c,d} such that the product of any two of them increased by 1 is a square — Part 2

$1 + a \,b = A^2$   …………………   $1 + b \,c = D^2$
$1 + a \,c = B^2$   …………………   $1 + b \,d = E^2$
$1 + a \,d = C^2$   …………………   $1 + c \,d = F^2$

We set   $a = 2$

$a = 2$
$b = 2 \,(n^2 - n)$
$c = 2 \, n^2 + 2 \, n$
$d = 4 \, (8 \, n^4 - 6 \, n^2 + 1)$

$1 + a \,b = (2 \, n - 1)^2$   ………………………   $1 + b \,c = (2 \, n^2 - 1)^2$
$1 + a \,c = (2 \, n + 1)^2$   ………………………   $1 + b \,d = (8 \, n^3 - 4 \, n^2 - 4 \, n + 1)^2$
$1 + a \,d = (8 \, n^2 - 3)^2$   …………………….   $1 + c \,d = (8 \, n^3 + 4 \, n^2 - 4 \, n - 1)^2$

## Set {a,b,c,d} such that the product of any two of them increased by 1 is a square — Part 1

$1 + a \,b = A^2$   ………..   $1 + b \,c = D^2$
$1 + a \,c = B^2$   ………..   $1 + b \,d = E^2$
$1 + a \,d = C^2$   ………..   $1 + c \,d = F^2$

In part 1,   we set   $a = 1$

$a = 1$
$b = n^2 - 1$
$c = n^2 + 2 \, n$
$d = 4 \, n^4 + 8 \, n^3 - 4 \, n$

$1 + a \,b = n^2$   ………………………………   $1 + b \,c = (n^2 + n - 1)^2$
$1 + a \,c = (n + 1)^2$   ………………………   $1 + b \,d = (2 \, n^3 + 2 \, n^2 - 2 \, n - 1)^2$
$1 + a \,d = (2 \, n^2 + 2 \, n - 1)^2$   ………….   $1 + c \,d = (2 \, n^3 + 4 \, n^2 - 1)^2$