Question

if n be any positive integer, show that the expression 5^5n - 3^4n is divisible by 4 and 11​

Answer 1

Answer:

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First proof:

We have:

n5–5n3+4n=n(n4–5n2+4)=n(n−2)(n−1)(n+1)(n+2)…(1)

Now, by (1) we see that:

n5–5n3+4n=(n−2)(n−1)n(n+1)(n+2)…(2)

The RHS of (2) is a product of five consecutine integers. Therefore, this product is divisible by 5 and since the product (n−1)n(n+1)(n+2) is formed by four consecutive integers, it follows that the RHS of (2) is also divisible by 4 . Therefore, we conclude that:

n5–5n3+4n=20m

where m is nonnegative integer.

Second proof:

We have:

n5–5n3+4n=(n5–n)−5n3+5n=(n5−n)−5(n3−n)...(1)

Now, we see that:

n5−n=n(n4−1)=n(n2−1)(n2+1)=n(n−1)(n+1)(n2+1)…(2)

For n≡0mod5=>n(n−1)(n+1)(n2+1)≡0mod5…(3)

For n≡1mod5=>(n−1)≡0mod5=>n(n−1)(n+1)(n2+1)≡0mod5…(4)

For n≡2mod5=>n2+1≡0mod5=>n(n−1)(n+1)(n2+1)≡0mod5…(5)

For n≡3mod5=>n2+1≡0mod5=>n(n−1)(n+1)(n2+1)≡0mod5…(6)

For n≡4mod5=>(n+1)≡0mod5=>n(n−1)(n+1)(n2+1)≡0mod5…(7)

Therefore, by (2) , (3) , (4) , (5) , (6) and (7) , it follows that:

n5−n≡0mod5...(8)

Hence, by (1) and (8) , we conclude that n5–5n3+4n is divisible by 5 .

Moreover by (1) , we obtain:

n5–5n3+4n=(n5−n)−5(n3−n)=n(n4−1)−5n(n2−1)=

n(n2−1)(n2+1)−5n(n2−1)=n(n2−1)(n2+1−5)=

n(n2−1)(n2−4)...(9)

Now, we discriminate two cases:

Case 1, n is even:

Then,there exists positive integer p such that n=2p , hence:

n(n2−1)(n2−4)=2p[(2p)2−1][(2p)2−4]=

2p[(2p)2–1]4(p2–1)≡0mod4

Case 2, n is odd:

Then,there exists positive integer p such that n=2p+1 , hence:

n(n2−1)(n2−4)=(2p+1)[(2p+1)2−1][(2p+1)2−4]=

(2p+1)[4p2+4p+1−1][(2p+1)2−4]=

(2p+1)[4(p2+p)][(2p+1)2−4]≡0mod4

Therefore, we have proved that:

n5–5n3+4n≡0mod20

Third proof:

In order to show that the polynomial n5–5n3+4n is divisible by 20 , it is enough to show that it is divisible by 4 and 5 .

Divisibility by 5:

We have:

n5–5n3+4n=(n5+4n)−5n3=n(n4+4)−5n…(1)

Hence, since 5n is divisible by 5 , it is enough to show that n(n4+4) is also divisible by 5 .

For n≡0mod5=>n(n4+4)≡0mod5...(2)

For n≡1mod5=>n4+4≡0mod5=>n(n4+4)≡0mod5…(3)

For n≡2mod5=>n4+4≡0mod5=>n(n4+4)≡0mod5…(4)

For n≡3mod5=>n4+4≡0mod5=>n(n4+4)≡0mod5…(5)

For n≡4mod5=>n4+4≡0mod5=>n(n4+4)≡0mod5…(6)

Therefore, by (2) , (3) , (4) , (5) and (6) , it follows that:

n(n4+4)≡0mod5...(7)

Divisibility by 4:

Now, since 4n is divisible by 4 , it is enough to show that n5−5n3 is also divisible by 4 .

We have:

n5−5n3=n3(n2−5)...(8)

For n≡0mod4=>n3(n2−5)≡0mod4...(9)

For n≡1mod4=>n2−5≡0mod4=>n3(n2−5)≡0mod4…(10)

For n≡2mod4=>n3≡0mod4=>n3(n2−5)≡0mod4…(11)

For n≡3mod4=>n2−5≡0mod4=>n3(n2−5)≡0mod4…(12)

herefore, by (9) , (10) , (11) and (12) , it follows that:

n5−5n3≡0mod4...(13)

Therefore, by (1) , (7) and (13) , we conclude that:

n5–5n3+4n≡0mod20