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Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?

Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?

BT.. Eat Your Heart Out

If the last four digits of my phone number are placed in front of the remaining three you get one more than twice my number! What is it?

Upsetting Pitagoras

Age 14 to 18
Challenge Level

"Hello. I'm Kim Jinhyuna from Kingston-Grammar School. I would like to inform you that I have worked out the question 'Upsetting Pitagoras'. The mathematical problem has an infinite number of solutions. Let $x=1$ and $y=2$. Then $${1\over x^2}+{1\over y^2}= {1\over 1^2}+{1\over 2^2} = 1{\cdot}25 = {1\over 0{\cdot}8}.$$ So $z^2 = 0{\cdot}8$, and $z = \sqrt{0{\cdot}8}$ which gives $z=\pm 0{\cdot}894$ to 3 significant figures. For another solution, let Let $x=3$ and $y=4$. Then $${1\over x^2}+{1\over y^2}= {1\over 3^2}+{1\over 4^2} = 1{\cdot}736\ldots = {1\over 5{\cdot}76}.$$ So $z^2 = 5{\cdot}76$, and $z = \sqrt{5{\cdot}76}$ which gives $z=\pm 2{\cdot}4$ (exactly). However, I think that the problem comes with the assumption that $x$, $y$ and $z$ are all integers, in which case one answer is $x=30$, $y=40$ and $z=24$; that is $${1\over 30^2}+{1\over 40^2} = {1\over 24^2}.$$ If we multiply both sides by $4$ we get $${1\over 15^2}+{1\over 20^2} = {1\over 12^2}.$$ I'm looking forward to more tough and hard questions."

You can also use Pythagorean Triples to find the smallest integer solution to the equation:

If $a^2 + b^2 = c^2$ then $${1\over b^2c^2} + {1\over a^2c^2} = {1\over a^2b^2}.$$ So every Pythgorean triple gives rise to a solution to our problem. The smallest solution of this type arises from $a=3, b=4, c=5$ and gives $${1\over 20^2} + {1\over 15^2} = {1\over 12^2}.$$ Notice we have not proved that there is no other way of producing solutions but a simple computer program to make an exhaustive check of values of $x, y$ and $z$ up to these values will prove that there are no smaller solutions.