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I prove using strong induction that for any n>5 P(n): n=4x+3y holds.
n=6: P(6): 6=4*0+3*2, thus x=0 and y=2
n=7: P(7): 7=4*1+3*1, thus x=y=1
I took two base cases because we’ll need to go two steps backwards in the induction step (k-2).
We assume that P(6) P(7) ……. P(k) is true for a positive integer k greater than or equal to 8.
We must prove that P(k+1) also holds.
Since k is greater than or equal to 8 => 5<k-2 < k, thus P(k-2) holds, so we can apply the induction hypothesis.
So, k-2=4x+3y => k+1=(k-2)+3=4x+3y+3=4x+3(y+1) which proves that k+1 has the same form.
We have proved that P(6) P(7) ……. P(k-2) => P(k+1)
According to strong induction principle, the proof is now complete and P(n) holds for any n>5.