ab + 5ac - 5bc - 25c² =7 a,b,c ∈ Z intregi
a ( b + 5c ) - 5c( b + 5c ) = 7
( a - 5c) · ( b + 5c) = 7 = 1 · 7
⇒ a - 5c =1
b + 5c = 7
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a - 5c + b + 5c = 1 +7 ⇒ a + b = 8
sau ( a - 5c ) · ( b + 5c) = ( -7) · ( -1 )
a - 5c = - 7
b + 5c = - 1
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a - 5c + b + 5c = -7 -1 ; a +b = - 8
I a +b i = I - 8 I = 8