Pell Equation
Nguyen Tuan Anh
14/03/2017 at 14:16
Thao Dola 14/03/2017 at 14:23
The first such triple is 8 = \(2^2+2^2\),9 = \(3^3+0^2\),10=\(3^2+1^2\), which suggests we consider triples \(x^21,x^2,x^2+1\).Since \(x^22y^2=1\) has infinitely many positive solutions (x,y), we see that \(x^21=y^2+y^2,x^2=x^2+0^2\)and \(x^2+1\) satisfy the requiment and there are infinitely many such triples.
Selected by MathYouLike 
FA KAKALOTS 28/01/2018 at 22:12
The first such triple is 8 = 22+22,9 = 33+02,10=32+12, which suggests we consider triples x2−1,x2,x2+1.Since x2−2y2=1 has infinitely many positive solutions (x,y), we see that x2−1=y2+y2,x2=x2+02and x2+1 satisfy the requiment and there are infinitely many such triples.

Such doge 14/03/2017 at 21:03
Wowe it hard
4799
questionsWeekly ranking
Tags
games 18
double counting 8
generating functions 2
probabilistic method 1
Polynomial 9
inequality 13
area 17
Equation 9
Primitive Roots Modulo Primes 1
Primitive in Arithmetic Progression 6
Base n Representatioons 4
Pell Equation 1
mixed number 1
Fraction 29
Circle 3
Imaginary numbers 1
Decimal number 2
Volume 2
percentages 6
simple equation 19
absolute value 19
rational numbers 20
Operation of Indices 21
Simulataneous Equation A System of Equations 25
Multiplication of Polynomial 17
divisibility 24
Maximum 5
Minimum 8
Fraction 4
Prime Numbers and Composite Numbers 13
Square Number 26
Even and Odd Numbers 13
GDC and LCM 11
GCD and LCM 12
Permutation and combination 9
combinations 5
interger 7
number 10
Diophantine equations 2
equations 1
Grade 6 19
Power 3
equality 2
maximaminima 2
square root 1
Polygon 2
IGCSE 1
factorial 1
integers 2