2n-1 | 7n+10 |*2
2n-1 | 2n-1 |*7
scazind cele e ecuatii:
2n-1 | 20+7=27
2n-1 | 27
Divizorii lui 27 sint: 1, 3, 9, 27, -1, -3, -9, -27
Sau 2n-1 ∈ {1, -1, 3, -3, 9, -9, 27, -27}
2n ∈ {2, 0, 4, -2, 10, -8, 28, -26}
n ∈ {1, 0, 2, -1, 5, -4, 14, -13}
Dar n ∈ N, deci n ∈ {0, 1, 2, 5, 14}
