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Diophantine N-tuples

Can you explain why a sequence of operations always gives you perfect squares?

DOTS Division

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}.

Sixational

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.

Lower Bound

Age 14 to 16
Challenge Level

Vassil Vassilev , a 14 year old Bulgarian student from St Michael's College, Leeds, sent the following solution:

Element No 1 $$ {1\over 2} + {2\over 1} = {1 + 4\over 2} = {5\over 2} = 2 + {1\over 2} = 2 + {1\over 1\times(1+1)} $$ Element No 2 $$ {2\over 3} + {3\over 2} = {4 + 9\over 6} = {13\over 6} = 2 + {1\over 6} = 2 + {1\over 2\times(2+1)} $$ Element No 3 $$ {3\over 4} + {4\over 3} = {9 + 16\over 12} = {25\over 12} = 2 + {1\over 12} = 2 + {1\over 3\times(3 + 1)} $$ Element No 4 $$ {4\over 5} + {5\over 4} = {16 + 15\over 20} = {41\over 20} = 2 + {1\over 20} = 2 + {1\over 4\times(4 + 1)} $$ Element No n

\begin{eqnarray} {n\over n + 1} + {n+1\over n} &=& {n^2 + (n + 1)^2\over n(n + 1)} = {n^2 + n^2 + 2n + 1\over n^2 + n}\\ &=& {2n^2 + 2n\over n^2 + n} + {1\over n^2 + n} = 2 + {1\over n^2 + n}\\ &=& 2 + {1\over n(n + 1)} \end{eqnarray}
Vassil stopped here but do you notice that, as $n$ gets bigger and bigger, so $$ {1\over n(n + 1)}$$ gets smaller and smaller and closer and closer to zero? If this sequence went on for ever the terms would get closer and closer to 2 without ever actually being equal to 2. Another way of seeing this is to look at $$ {n\over n + 1} $$ and see that this must get closer and closer to the value 1 as $n$ gets bigger and bigger and also $$ {n + 1\over n} $$ must also get closer and closer to 1, so the $n$th term of the sequence, namely $$ {n\over n + 1} + {n + 1\over n} $$ must get closer and closer to 2. We say the limit of this sequence, as $n$ tends to infinity, is 2.

Some, but not all, of the points on the graph of $$ y = x + {1\over x} $$ represent terms of the sequence. You might like to draw the graph and look at what happens to it around $x = 1$.

Vassil worked out the $n$th term of the similar sequence formed by adding the squares of these fractions and after doing some algebra to simplify the expression he obtained the following result: $$ \left({n\over n + 1}\right)^2 + \left({n + 1\over n}\right)^2 = 2 + \left({2n + 1\over n(n + 1)}\right)^2. $$ Here again the limit as n tends to infinity is 2. You might like to prove this result in another way.