### Perfect square problems

Number pattern

1 ´ 2 ´ 3 ´ 4 + 1                  = 25       = 52

2 ´ 3 ´ 4 ´ 5 + 1          = 121      = 112

3 ´ 4 ´ 5 ´ 6 + 1          = 361      = 192

5 ´ 6 ´ 7 ´ 8 + 1          = 841        = 292

:        :        :        :        :

Algebra

The pattern seems to work onwards. Can we prove this?

Simple! Use algebra!

n(n + 1)(n + 2)(n + 3) + 1

= [n(n + 3)] [(n + 1)(n + 2)] + 1

= [n2 + 3n] [n2 + 3n + 2] + 1

= [(n2 + 3n + 1) – 1] [(n2 + 3n + 1) + 1] + 1

= (n2 + 3n + 1)2 – 12 + 1              [since  ( x – y )(x + y) = x2 – y2]

= (n2 + 3n + 1)2,         which is a complete square.

\ n(n + 1)(n + 2)(n + 3) + 1 = (n2 + 3n + 1)2

If you put in n = 1, 2, 3, 4, ….  , you can get the number pattern on the top.

Number pattern

49                     =      72

4489                =      672

444889           =      6672

44448889      =      66672

:        :        :        :

Algebra

Let  m = 11….1 (n times)

\ 10n  =  9 × 11….1   + 1 = 9m + 1

(n times)

44….……..488……..…..89     =     44….…4 ×10n + 88…..…8 + 1

(n times)         (n-1 times)                      (n times)                  (n times)

=      4m (9m + 1) + 8m + 1

=     36 m2 + 4m + 8 m + 1

=     36 m2 + 12m + 1

=     (6m + 1)2

=      (66………6 + 1)2

(n times)

=     (66…..…..67)2

(n-1 times)

Number pattern

11                    – 2                   =      9               =      32

1111                – 22                =     1089               =     332

111111           – 222              =      110889            =     3332

11111111       – 2222           =      11108889       =      33332

:        :        :        :        :

Algebra

Let        m = 11….11 (n times)

Then       22….22 (n times)        =     2m

11….11  (2n times)      =      11….11 (n times) ´ 100….00 (n zeros) + 11….11 (n times)

=     100….00 (n zeros) ´ m + m

\     11….11 (2n times) – 22….22 (n times)

=     [100….00 (n zeros) ´ m + m] – 2m

=      100….00 (n zeros) ´ m – m

=      99….99 (n times) ´ m

=      9m ´ m

=      (3m)2

=      (33….33)2  (n threes)