Integers(x,y,z) such that (xy+x+y),(yz+y+z),(zx+z+x) are all squares

 
 
Book III Diophantus Arithmetica:

To find integers   (x, y, z)   such that

x y \; + \; x \; + \; y
y z \; + \; y \; + \; z
z x \; + \; z \; + \; x
x y \; + \; z
x z \; + \; y
y z \; + \; x

are all squares

Solution:

x \; = \; n^2,          y \; = \; (n + 1)^2,          z \; = \; 4(n^2 + n + 1)

Indeed,

x y \; + \; x \; + \; y
= \;  n^2((n + 1)^2) \; + \; n^2 \; + \; (n + 1)^2
= \;  n^4+2 n^3+3 n^2+2 n+1
= \;  (n^2+n+1)^2

y z \; + \; y \; + \; z
= \;  (n + 1)^2(4(n^2 + n + 1)) \; + \; (n + 1)^2 \; + \; 4(n^2 + n + 1)
= \;  4 n^4+12 n^3+21 n^2+18 n+9
= \;  (2 n^2+3 n+3)^2

z x \; + \; z \; + \; x
= \;  (4(n^2 + n + 1))n^2 + 4(n^2 + n + 1) \; + \; n^2
= \;  4 n^4+4 n^3+9 n^2+4 n+4
= \;  (2 n^2+n+2)^2

x y \; + \; z
= \; n^2(n + 1)^2 \; + \; 4(n^2 + n + 1)
= \; n^4+2 n^3+5 n^2+4 n+4
= \; (n^2+n+2)^2

x z \; + \; y
= \; n^2(4(n^2 + n + 1)) \; + \; (n + 1)^2
= \; 4 n^4+4 n^3+5 n^2+2 n+1
= \; (2 n^2+n+1)^2

y z \; + \; x
= \; (n + 1)^2(4(n^2 + n + 1)) \; + \; n^2
= \; 4 n^4+12 n^3+17 n^2+12 n+4
= \; (2 n^2+3 n+2)^2

x y \; + \; y z \; + \; z x
= \; n^2(n + 1)^2 \; + \; (n + 1)^2(4(n^2 + n + 1)) \; + \; n^2(4(n^2 + n + 1))
= \; 9 n^4+18 n^3+21 n^2+12 n+4
= \; (3 n^2+3 n+2)^2

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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