Exista d divide 7n+8 inmultim cu 4=>d divide 28n+32
d divide 4n+5 inmultim cu 7=> d divide 28n+35, le scadem si =>
d divide 3,2x3,3x3,4x3,5x3,6x3....
din 4n+5=3x3=>n=1
4n+5=7x3=>4n=21-5, n=4
4n+5=11x3=> 4n=33-5, n=7
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n={1,4,7,10,13,16,19}