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# Janine's Conjecture

(*x* - 1) *x* ( *x* + 1) + *x* =
*x* ^{3} .

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Julia of Downe House School gave the neatest solution to this problem by substituting 'x-1', 'x', 'x+1' for the three consecutive numbers and giving the following statement of Janine's conjecture:

(

This is Julia's proof:

( *x* - 1) ( *x* + 1) = *x* ^{2} -
1

and

( *x* ^{2} - 1) *x* = *x* ^{3}
- *x* .

Therefore ( *x* - 1) *x* ( *x* + 1) +
*x* = *x* ^{3} .

So Janine's conjecture will always work whichever three consecutive
numbers are chosen.

Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.

The nth term of a sequence is given by the formula n^3 + 11n. Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.