# Minus One But What a Difference

Prove that for each integer n ≥ 3 the equation

x^{ n} + y^{ n} = z^{ n-1}

has infinitely many solutions in positive integers.

### References

- T. Andreescu, D. Andrica, I. Cucurezeanu,
*An Introduction to Diophantine Equations*, Birkhäuser, 2010, p. 23

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

Just verify that

x = k(k

y = (k

z = (k

^{n}+ 1)^{n-2},y = (k

^{n}+ 1)^{n-2}, andz = (k

^{n}+ 1)^{n-1}solve the equation. Indeed,

[k(k

^{n}+ 1)^{n-2}]^{n}+ [(k^{n}+ 1)^{n-2}]^{n}= (k^{n}+ 1)^{n(n-2) + 1}= (k^{n}+ 1)^{(n-1)²}while also

[(k

^{n}+ 1)^{n-1}]^{n-1}= (k^{n}+ 1)^{(n-1)²}.Why does not this work for the FLT? Just minus 1 - but what a difference!

An even stranger example has been furnished by Emmanuel Moreau from France. Set

x = k

y = x,

z = k

^{n-1}· 2^{n-2},y = x,

z = k

^{n}· 2^{n-1}.and verify that x^{n} + y^{n} = z^{n-1}. For the original FLT equation one would not dream of having

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

71948605