Math  /  Algebra

Question10. For n1n \geq 1, establish that the integer n(7n2+5)n\left(7 n^{2}+5\right) is of the form 6k6 k.
11. If nn is an odd integer, show that n4+4n2+11n^{4}+4 n^{2}+11 is of the form 16k16 k.

Studdy Solution

STEP 1

1. We are given two separate problems to solve.
2. For problem 10, we need to show that the expression n(7n2+5) n(7n^2 + 5) is divisible by 6 for n1 n \geq 1 .
3. For problem 11, we need to show that the expression n4+4n2+11 n^4 + 4n^2 + 11 is divisible by 16 when n n is an odd integer.

_HIGH_LEVEL_APPROACH_ for Problem 10:
1. Analyze the expression n(7n2+5) n(7n^2 + 5) for divisibility by 2.
2. Analyze the expression n(7n2+5) n(7n^2 + 5) for divisibility by 3.
3. Conclude that n(7n2+5) n(7n^2 + 5) is divisible by 6.

_HIGH_LEVEL_APPROACH_ for Problem 11:
1. Substitute n=2m+1 n = 2m + 1 to express n n as an odd integer.
2. Expand and simplify the expression n4+4n2+11 n^4 + 4n^2 + 11 .
3. Analyze the simplified expression for divisibility by 16.

**Problem 10:**

STEP 2

STEP 3

Consider the expression n(7n2+5) n(7n^2 + 5) .
For divisibility by 2: - If n n is even, n n is divisible by 2. - If n n is odd, then 7n2 7n^2 is odd and 7n2+5 7n^2 + 5 is even, hence divisible by 2.

STEP 4

For divisibility by 3: - Consider n0,1,2(mod3) n \equiv 0, 1, 2 \pmod{3} .
Case 1: n0(mod3) n \equiv 0 \pmod{3} - n(7n2+5)0(mod3) n(7n^2 + 5) \equiv 0 \pmod{3} .
Case 2: n1(mod3) n \equiv 1 \pmod{3} - 7n2+57(1)2+5120(mod3) 7n^2 + 5 \equiv 7(1)^2 + 5 \equiv 12 \equiv 0 \pmod{3} .
Case 3: n2(mod3) n \equiv 2 \pmod{3} - 7n2+57(4)+5330(mod3) 7n^2 + 5 \equiv 7(4) + 5 \equiv 33 \equiv 0 \pmod{3} .

STEP 5

Since n(7n2+5) n(7n^2 + 5) is divisible by both 2 and 3, it is divisible by 6.
**Problem 11:**
STEP_1: Let n=2m+1 n = 2m + 1 where m m is an integer, since n n is odd.
Substitute into the expression:
n4+4n2+11=(2m+1)4+4(2m+1)2+11 n^4 + 4n^2 + 11 = (2m + 1)^4 + 4(2m + 1)^2 + 11
STEP_2: Expand and simplify:
(2m+1)4=16m4+32m3+24m2+8m+1 (2m + 1)^4 = 16m^4 + 32m^3 + 24m^2 + 8m + 1 4(2m+1)2=16m2+16m+4 4(2m + 1)^2 = 16m^2 + 16m + 4
Combine terms:
n4+4n2+11=16m4+32m3+40m2+24m+16 n^4 + 4n^2 + 11 = 16m^4 + 32m^3 + 40m^2 + 24m + 16
STEP_3: Notice that each term is divisible by 16:
n4+4n2+11=16(m4+2m3+2.5m2+1.5m+1) n^4 + 4n^2 + 11 = 16(m^4 + 2m^3 + 2.5m^2 + 1.5m + 1)
Since m4+2m3+2.5m2+1.5m+1 m^4 + 2m^3 + 2.5m^2 + 1.5m + 1 is an integer, n4+4n2+11 n^4 + 4n^2 + 11 is of the form 16k 16k .
Both problems are solved, confirming the divisibility conditions.

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