If ( a 2 + a b + b 2 ) ( c 2 + c d + d 2 ) = e 2 + e f + f 2 , and e = m 1 a c + m 2 a d + m 3 b c + m 4 b d and f = n 1 a c + n 2 a d + n 3 b c + n 4 b d , find the number of integer solutions for ( m 1 , m 2 , m 3 , m 4 , n 1 , n 2 , n 3 , n 4 ) .
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Expanding ( a 2 + a b + b 2 ) ( c 2 + c d + d 2 ) gives a 2 c 2 + a 2 d 2 + b 2 c 2 + b 2 d 2 + a b c 2 + a 2 c d + b 2 c d + a b d 2 + a b c d .
Expanding e 2 + e f + f 2 gives ( m 1 2 + m 1 n 1 + n 1 2 ) a 2 c 2 + ( m 2 2 + m 2 n 2 + n 2 2 ) a 2 d 2 + ( m 3 2 + m 3 n 3 + n 3 2 ) b 2 c 2 + ( m 4 2 + m 4 n 4 + n 4 2 ) b 2 d 2 + ( 2 m 1 m 3 + m 3 n 1 + m 1 n 3 + 2 n 1 n 3 ) a b c 2 + ( 2 m 1 m 2 + m 2 n 1 + m 1 n 2 + 2 n 1 n 2 ) a 2 c d + ( 2 m 3 m 4 + m 3 n 4 + m 4 n 3 + 2 n 3 n 4 ) b 2 c d + ( 2 m 2 m 4 + m 2 n 4 + m 4 n 2 + 2 n 2 n 4 ) a b d 2 + ( 2 m 1 m 4 + 2 m 2 m 3 + m 1 n 4 + m 2 n 3 + m 3 n 2 + m 4 n 1 + 2 n 1 n 4 + 2 n 2 n 3 ) a b c d
For the a 2 c 2 term we have 1 = m 1 2 + m 1 n 1 + n 1 2
For the a 2 d 2 term we have 1 = m 2 2 + m 2 n 2 + n 2 2
For the b 2 c 2 term we have 1 = m 3 2 + m 3 n 3 + n 3 2
For the b 2 d 2 term we have 1 = m 4 2 + m 4 n 4 + n 4 2
For the a b c 2 term we have 1 = 2 m 1 m 3 + m 3 n 1 + m 1 n 3 + 2 n 1 n 3
For the a 2 c d term we have 1 = 2 m 1 m 2 + m 2 n 1 + m 1 n 2 + 2 n 1 n 2
For the b 2 c d term we have 1 = 2 m 3 m 4 + m 3 n 4 + m 4 n 3 + 2 n 3 n 4
For the a b d 2 term we have 1 = 2 m 2 m 4 + m 2 n 4 + m 4 n 2 + 2 n 2 n 4
For the a b c d term we have 1 = 2 m 1 m 4 + 2 m 2 m 3 + m 1 n 4 + m 2 n 3 + m 3 n 2 + m 4 n 1 + 2 n 1 n 4 + 2 n 2 n 3
For x 2 + x y + y 2 = 1 to have real solutions, its discriminant (in terms of x ) 1 − 4 ( y 2 − 1 ) > 0 solves to y 2 < 4 5 . If y is an integer, it is limited to y = ± 1 or y = 0 . Likewise by symmetry, if x is an integer, it is limited to x = ± 1 or x = 0 . Since from above we know that 1 = m 1 2 + m 1 n 1 + n 1 2 , 1 = m 2 2 + m 2 n 2 + n 2 2 , 1 = m 3 2 + m 3 n 3 + n 3 2 , and 1 = m 4 2 + m 4 n 4 + n 4 2 , we know that all m and n are in the same form as x and y and their values must also be limited to ± 1 or 0 .
With these limiting values of m and n and a little bit of algebraic effort with the above equations, we find the following integer solutions for ( m 1 , m 2 , m 3 , m 4 , n 1 , n 2 , n 3 , n 4 ) :
( − 1 , − 1 , − 1 , 0 , 0 , 1 , 1 , 1 ) , ( − 1 , − 1 , − 1 , 0 , 1 , 0 , 0 , − 1 ) , ( − 1 , − 1 , 0 , − 1 , 0 , 1 , − 1 , 0 ) , ( − 1 , − 1 , 0 , − 1 , 1 , 0 , 1 , 1 ) ,
( − 1 , 0 , − 1 , − 1 , 0 , − 1 , 1 , 0 ) , ( − 1 , 0 , − 1 , − 1 , 1 , 1 , 0 , 1 ) , ( − 1 , 0 , 0 , 1 , 0 , − 1 , − 1 , − 1 ) , ( − 1 , 0 , 0 , 1 , 1 , 1 , 1 , 0 ) ,
( 0 , − 1 , − 1 , − 1 , − 1 , 0 , 0 , 1 ) , ( 0 , − 1 , − 1 , − 1 , 1 , 1 , 1 , 0 ) , ( 0 , − 1 , 1 , 0 , − 1 , 0 , − 1 , − 1 ) , ( 0 , − 1 , 1 , 0 , 1 , 1 , 0 , 1 ) ,
( 0 , 1 , − 1 , 0 , − 1 , − 1 , 0 , − 1 ) , ( 0 , 1 , − 1 , 0 , 1 , 0 , 1 , 1 ) , ( 0 , 1 , 1 , 1 , − 1 , − 1 , − 1 , 0 ) , ( 0 , 1 , 1 , 1 , 1 , 0 , 0 , − 1 ) ,
( 1 , 0 , 0 , − 1 , − 1 , − 1 , − 1 , 0 ) , ( 1 , 0 , 0 , − 1 , 0 , 1 , 1 , 1 ) , ( 1 , 0 , 1 , 1 , − 1 , − 1 , 0 , − 1 ) , ( 1 , 0 , 1 , 1 , 0 , 1 , − 1 , 0 ) ,
( 1 , 1 , 0 , 1 , − 1 , 0 , − 1 , − 1 ) , ( 1 , 1 , 0 , 1 , 0 , − 1 , 1 , 0 ) , ( 1 , 1 , 1 , 0 , − 1 , 0 , 0 , 1 ) , ( 1 , 1 , 1 , 0 , 0 , − 1 , − 1 , − 1 )
for a total of 2 4 solutions.