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Home / Elementary Number Theory / For n ≥ 1, established that the integer n(7n^2 + 5) is of the form 6k

For n ≥ 1, established that the integer n(7n^2 + 5) is of the form 6k

Solution: According to division algorithm, we have

n = 6q + r ; 0 ≤ r < 6

Let, A = n(7n2 + 5)

For r = 0, we have

A = n(7n2 + 5)

= (6q + r) {7(6q + r)2 + 5}

= 6q (7.36q2 + 5)

= 6(252q3 + 5q)

Let, k = 252q3 + 5q

∴ A = 6k

For r = 1, we have

A = n(7n2 + 5)

= (6q + r) {7(6q + r)2 + 5}

= (6q + 1) {7(6q + 1)2 + 5}

= 7(6q + 1)3 + 5(6q + 1)

= 7(216q3 + 108q2 + 18q + 1) + 30q + 5

= 1512q3 + 756q2+126q+7 + 30q+5

= 1512q3 + 756q2 + 156q + 12

= 6(252q3 + 126q2 +26q + 2)

Let, k = 252q3 + 126q2 +26q + 2

∴ A = 6k

For r = 2, we have

A = n(7n2 + 5)

= (6q + r) {7(6q + r)2 + 5}

= (6q + 2) {7(6q + 2)2 + 5}

= 7(6q + 2)3 + 5(6q + 2)

= 7(216q3 + 216q2 + 72q + 8) + 30q + 10

= 1512q3 + 1512q2+504q+56 + 30q+10

= 1512q3 + 1512q2 + 534q + 66

= 6(252q3 + 252q2 +69q + 11)

Let, k = 252q3 + 252q2 +69q + 11

∴ A = 6k

For r = 3, we have

A = n(7n2 + 5)

= (6q + r) {7(6q + r)2 + 5}

= (6q + 3) {7(6q + 3)2 + 5}

= 7(6q + 3)3 + 5(6q + 3)

= 7(216q3 + 324q2 + 162q + 27) + 30q + 15

= 1512q3 + 2268q2+1134q+189 + 30q+15

= 1512q3 + 2268q2 + 1134q + 204

= 6(252q3 + 378q2 +244q + 34)

Let, k = 252q3 + 378q2 +244q + 34

∴ A = 6k

For r = 4, we have

A = n(7n2 + 5)

= (6q + r) {7(6q + r)2 + 5}

= (6q + 4) {7(6q + 4)2 + 5}

= 7(6q + 4)3 + 5(6q + 4)

= 7(216q3 + 432q2 + 288q + 64) + 30q + 20

= 1512q3 + 3024q2+2016q+488 + 30q+20

= 1512q3 + 3024q2 + 2046q + 468

= 6(252q3 + 504q2 +341q + 78)

Let, k = 252q3 + 504q2 +341q + 78

∴ A = 6k

For r = 5, we have

A = n(7n2 + 5)

= (6q + r) {7(6q + r)2 + 5}

= (6q + 5) {7(6q + 5)2 + 5}

= 7(6q + 5)3 + 5(6q + 5)

= 7(216q3 + 540q2 + 450q + 125) + 30q + 25

= 1512q3 + 3780q2+3150q+875+ 30q+25

= 1512q3 + 3780q2 + 3180q + 900

= 6(252q3 + 630q2 +530q + 150)

Let, k = 252q3 + 630q2 +530q + 150

∴ A = 6k

So that, For n ≥ 1, established that the integer n(7n2 + 5) is of the form 6k. (Proved)

Problem: If n is an odd integer, show that n4 + 4n2 + 11 is of the form 16k.

Solution: Let, n = 2q + 1

Now, n4 + 4n2 +11 = (n2 + 2.n2.2 + 4) + 7

= (n2 + 2)2 + 7

= {(2q + 1)2 + 2} + 7

= (4q2 + 4q + 1 + 2)2 + 7

= (4q2 + 4q + 3)2 + 7

= (4q2 + 4q)2 + 2 . (4q2 + 4q) . 3 + 9 +7

= 16q4 + 32q3 + 16q2 + 24q2 + 24q + 16

= 16q4 + 32q3 + 40q2 + 24q + 16

Let, q = 2r and 2r + 1

When q = 2r, then

16q4 + 32q3 + 40q2 + 24q + 16 = 16(2r)4 + 32(2r)3 + 40(2r)2 + 24(2r) + 16

= 16(2r)4 + 32(2r)3 + 160 r2 + 48 r + 16

= 16 {(2r)4 + 2(2r)3 + 10 r2 + 3 r + 1}

Let, k = {(2r)4 + 2(2r)3 + 10 r2 + 3 r + 1}

∴ 16q4 + 32q3 + 40q2 + 24q + 16 = 16k

When q = 2 r + 1, then

16q4 + 32q3 + 40q2 + 24q + 16

= 16(2r + 1)4 + 32(2r + 1)3 + 40(2r + 1)2 + 24(2r + 1) + 16

= 16(2r + 1)4 + 32(2r + 1)3 + 160 r2 + 208 r + 80

= 16 {(2r + 1)4 + 2(2r + 1)3 + 10 r2 + 13 r + 5}

Let, k = {(2r + 1)4 + 2(2r + 1)3 + 10 r2 + 13 r + 5}

∴ 16q4 + 32q3 + 40q2 + 24q + 16 = 16k

So that, If n is an odd integer then n4 + 4n2 + 11 is of the form 16k. (Proved)

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