Olympiad Proof Problem - Day 4

Find all primes such that p3q7=pqp^3 - q^7 = p-q.


Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
4 years ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Only one pair of primes exist. I used casework to solve this.

We have

p3q7=pqp^3-q^7=p-q

p3p=q7qp^3-p=q^7-q

p(p1)(p+1)=q(q31)(q3+1)p(p-1)(p+1)=q(q^3-1)(q^3+1)

p(p1)(p+1)=q(q1)(q+1)(q2+q+1)(q2q+1)(1)p(p-1)(p+1)=q(q-1)(q+1)(q^2+q+1)(q^2-q+1) \quad \ldots (1)

I will first check all cases when p3<q7p^3 < q^7, that is, when p3q7p^3-q^7 is negative. Note that since pp and qq are primes, one of qq, q1q-1, q+1q+1, q2+q+1q^2+q+1 or q2q+1q^2-q+1 is equal to pp assuming q2q \neq 2 (This issue will be addressed).

Since the difference is negative, we must have p<qp < q. Combining this with what we know, we have p=q1p=q-1. (Note: pq2q+1p \neq q^2 - q + 1 since the quadratic is greater than qq for all positive qq.) Since pp and qq are primes, we have p=2p=2, q=3q=3. Checking, we find that this is incorrect so there are no solutions in this case.

Now, we check when p=qp = q. We get that p1=q1p-1 = q-1 and p+1=q+1p+1 = q+1, rendering q2+q+1=1q^2+q+1=1 and q2q+1=1q^2-q+1=1, which gives p=q=0p=q=0 which are not primes. There are no solutions in this case.

Finally, we check when p>qp > q. We have the following values for pp: p=q+1p=q+1, p=q2q+1p=q^2-q+1, p=q2+q+1p=q^2+q+1. We will now look at each expresion for pp.

Case 1: p=q+1p=q+1

We get that p=3p=3, q=2q=2. Checking, we find that this is incorrect. Therefore, there are no solutions in this case.

Case 2: p=q2q+1p=q^2-q+1

We have p=q2q+1p=q^2-q+1 so p1=q2q=q(q1)p-1=q^2-q=q(q-1) which leaves p+1=(q+1)(q2+q+1)p+1=(q+1)(q^2+q+1) from (1)(1). We now have

q2q+1+1=(q+1)(q2+q+1)q^2-q+1+1=(q+1)(q^2+q+1)

q2q+2=q3+2q2+2q+1q^2-q+2=q^3+2q^2+2q+1

q3+q2+3q1=0q^3+q^2+3q-1=0

Playing with this cubic, we find that it has no integral solutions for qq (There are many different ways to check). Hence, there are no solutions in this case.

Case 3: p=q2+q+1p=q^2+q+1

In this case, we have p=q2+q+1p=q^2+q+1, so p1=q2+q=q(q+1)p-1=q^2+q=q(q+1), which leaves p+1=(q1)(q2q+1)p+1=(q-1)(q^2-q+1) from (1)(1). We now have the following:

q2+q+1+1=(q1)(q2q+1)q^2+q+1+1=(q-1)(q^2-q+1)

q2+q+2=q32q2+2q1q^2+q+2=q^3-2q^2+2q-1

q33q2+q3=0q^3-3q^2+q-3=0

(q2+3)(q3)=0(q^2+3)(q-3)=0

Clearly, only q=3q=3 is the integral solution in this case, which implies p=13p=13. Checking, we find that this is true. Therefore, only one solution exists in this case: p=13p=13, q=3q=3.

Thus, the only solution which satisfies is p=13p=13, q=3q=3.

Great problem, Surya.

Sharky Kesa - 3 years, 12 months ago

Log in to reply

This is a great problem and solution!

Sualeh Asif - 3 years, 12 months ago

Log in to reply

Same solution.

Abhi Kumbale - 2 years, 10 months ago

Log in to reply

@Surya Prakash , is this a satisfactory solution?

Sharky Kesa - 3 years, 12 months ago

Log in to reply

Respected Sharky, I did not understand the second sentence below equation 1 ( Note that since p & q are primes, ...) I request you to explain.

Պոոռնապռագնյա ՊՌ - 3 years, 12 months ago

Log in to reply

@Պոոռնապռագնյա Պռ Since pp and qq are primes, they both have 2 factors each: 11 and the number itself. This implies that pp is equivalent to one of the expressions in the brackets.

Sharky Kesa - 3 years, 12 months ago

Log in to reply

No such pair of primes exist.

Saakshi Singh - 4 years ago

Log in to reply

No. There exists.

Surya Prakash - 4 years ago

Log in to reply

Check out p=13 and q=3.

Sharky Kesa - 3 years, 12 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...