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## 'Shades of Fermat's Last Theorem' printed from http://nrich.maths.org/

The familiar Pythagorean relationship $3^2+ 4^2=5^2$ is 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 $5$ ? The
question asks you to prove that there are exactly three
solutions.

Both Alan Riddell of Madras
College, St Andrew's and Edward Wallace of Graveney School,
Tooting solved this problem. When $n = 2$ $$(x - 1)^n + x^n
= (x + 1)^n$$ becomes $$(x - 1)^2 + x^2 = (x + 1)^2$$ $$x^2 - 2x +1
+ x^2 = x^2 +2x +1$$ $$x^2 - 4x = x(x - 4) = 0$$ so $x = 0$ or $x =
4$. The three solutions to this problem are $x=4, n=2$ where $3^2 +
4^2 = 5^2;$ $x=0, n=2$ where $(-1)^2 +0^2 = 1^2$ and $x=0, n=4 $
where $(-1)^4 +0^4 = 1^4.$

We now show that there are no other solutions. For $n = 3$ we
seek solutions of $$(x - 1)^3 + x^3 = (x + 1)^3$$ Simplifying:
$$x^3 -3x^2 +3x -1 +x^3 = x^3 +3x^3 +3x +1$$ $$x^3 -6x^2 = 2 $$
$$x^2(x - 6) = 2.$$ If any solution exists then $x> 6$ because
the right hand side is positive so $x \geq 7 $ but then $x^2(x - 6)
\geq 49 $ so there are no solutions. For $n = 4$ we seek solutions
of $$(x - 1)^4 + x^4 = (x + 1)^4$$ Simplifying: $$x^4 - 4x^3 + 6x^2
- 4x +1 +x^4 = x^4 + 4x^3 + 6x^2 + 4x +1$$ $$x^4 - 8x^3 - 8x = 0$$
This gives the solution $x=0,n=4$ where $(-1)^4 +0^4 = 1^4.$ If
another solution exists for $n = 4$ then $x^2(x - 8) = 8$ so $x>
8$ because the right hand side is positive. In this case $x \geq 9$
but then $x^2(x - 8) \geq 81$ so there are no solutions. For $n =
5$ we seek solutions of $$(x - 1)^5+ x^5 = (x + 1)^5$$ Simplifying:
$$x^5 - 10x^4 -20x^2 - 2 = 0$$ If a solution exists for $n = 5$
then $x^2(x^3 - 10x^2 - 20) = 2$. Clearly this has no solutions
because both sides are positive so $x > 10$ but then $x^2(x^3 -
10x^2 - 20) > 100$ and cannot equal 2.