# SOLUTION OF MATH 307 MIDTERM 1

# OLUTION OF MATH 307 MIDTERM 1

Note that for solutions of numbers 3 (a), (b), and 4, 5, I
gave necessary hints

for you only, not the complete solutions as students should show in their work.

**1. The Mathematical Induction
**(a) (4 points) Use the mathematical induction to prove the formula:

**Solution. **For n = 1 : 1^{2} = 1, and
,
correct.

Assume that the formula is correct for n = k ≥ 1, that means for this k,

Now we prove the formula for n = k + 1. Consider 1^{2} + 2^{2}
+ 3^{2} + · · · + k^{2} + (k +

1)^{2}

Compare
with the given

formula we see that it is correct for n = k + 1. Hence the formula is proved.

(b) (5 points) Use the mathematical induction to prove
that for n > 0, 15 is a

divisor of ().

**Solution.** For n = 1,,
and 15|15, correct.

Assume that the statement is correct for n = k ≥ 1, that means
.
For

n = k+1 we see

.
But by our induction assumption,
is divisible by 15. Hence

is divisible by 15. The proof is complete.

**2. The Binomial Theorem
**(a) (3 points) Write the binomial formula: (a + b)

^{n}

**Solution**.

(b) (4 points) Prove that

**Solution.** Using (a) for a = b = 1.

**3. Prime numbers
**(a) (5 points) Let n > 1 be an integer. Prove that
is not a prime number.

Hint:

**Solution:**
for n > 1 we see each

factor is greater than 1. Hence
is NOT prime.

(b) (4 points) Show that for any integer n > 1, n^{5} + 1 is
not a prime number.

**Solution.** Use the formula n^{5} + 1 = (n + 1)(n^{4} −
n^{3} + n^{2} − n + 1). For n > 1

each factor is greater than 1.

(c) (6 points) Let p be a prime number. Find all values of
p such that 5p + 1 is

a perfect square .

**Solution.** Since 5p + 1 is a pefect square, 5p + 1 =
a^{2} for some positive integer

a. Hence 5p = a^{2} − 1 = (a − 1)(a + 1). Therefore
Now, since 5 is a

prime, we must have either 5|(a − 1) or 5|(a + 1). This together with the fact
that p is

a prime gives either a − 1 = 5 or a + 1 = 5. The first case yields a = 5 + 1 = 6
and so

p = 7. The second case gives a = 5− 1 = 4, and so p = 3.

Answer: p = 7 or p = 3.

**Remark.** From the previous part (c), we can state a
general problem: Let q, p

be prime numbers such that qp+1 is a perfect square. Find q and p. Using the
solution

of (c) above, you can try to answer this problem. A list of such q, p less than
106 was

found today (4/27) by my younger son (an undergraduate computer science student

at HTC, OU). There are 78498 prime numbers in the range [2, 1000000] in which
8169

prime-pairs that satisfy the condition pq + 1 = n^{2}. View attachment for
details.

**4. Greatest common divisor (gcd) and least common
multiple (lcm)
**(a) (4 points) Find gcd and lcm of 2030, and 4246.

**Solution.** Use Euclidian algorithm, or write these
numbers in the product of

primes to find the answers for gcd. For lcm use the formula

(b) (5 points) Find x and y such that gcd(48, 138) = 48x + 138y.

**Solution.** Use Euclidian Algorithm. Answer: x = 3, y
= −1

**5. Linear Diophantine Equations
**Find the general solution of the following equations:

(a) (5 points) 24x + 36y = 156.

**Solution.** Use Euclidian Algorithm to find gcd(36,
24) = 12 and a specific

solution x_{0} = −13, y_{0} = 13. Then the general solution is:

x = −13 + 3t, y = 13 − 2t for any t ∈R.

(b) (5 points) 102x + 1003y = 221.

**Solution.** Do the same way as part (a). Answer: x_{0}
= 130, y_{0} = −13,

gcd(102, 1003) = 17. Hence the general solution is: x = 130 + 59t, y = −13 − 6t

for all t ∈R.

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