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Only one pair of primes exist. I used casework to solve this.
We have
p3−q7=p−q
p3−p=q7−q
p(p−1)(p+1)=q(q3−1)(q3+1)
p(p−1)(p+1)=q(q−1)(q+1)(q2+q+1)(q2−q+1)…(1)
I will first check all cases when p3<q7, that is, when p3−q7 is negative. Note that since p and q are primes, one of q, q−1, q+1, q2+q+1 or q2−q+1 is equal to p assuming q=2 (This issue will be addressed).
Since the difference is negative, we must have p<q. Combining this with what we know, we have p=q−1. (Note: p=q2−q+1 since the quadratic is greater than q for all positive q.) Since p and q are primes, we have p=2, q=3. Checking, we find that this is incorrect so there are no solutions in this case.
Now, we check when p=q. We get that p−1=q−1 and p+1=q+1, rendering q2+q+1=1 and q2−q+1=1, which gives p=q=0 which are not primes. There are no solutions in this case.
Finally, we check when p>q. We have the following values for p: p=q+1, p=q2−q+1, p=q2+q+1. We will now look at each expresion for p.
Case 1:p=q+1
We get that p=3, q=2. Checking, we find that this is incorrect. Therefore, there are no solutions in this case.
Case 2:p=q2−q+1
We have p=q2−q+1 so p−1=q2−q=q(q−1) which leaves p+1=(q+1)(q2+q+1) from (1). We now have
q2−q+1+1=(q+1)(q2+q+1)
q2−q+2=q3+2q2+2q+1
q3+q2+3q−1=0
Playing with this cubic, we find that it has no integral solutions for q (There are many different ways to check). Hence, there are no solutions in this case.
Case 3:p=q2+q+1
In this case, we have p=q2+q+1, so p−1=q2+q=q(q+1), which leaves p+1=(q−1)(q2−q+1) from (1). We now have the following:
q2+q+1+1=(q−1)(q2−q+1)
q2+q+2=q3−2q2+2q−1
q3−3q2+q−3=0
(q2+3)(q−3)=0
Clearly, only q=3 is the integral solution in this case, which implies p=13. Checking, we find that this is true. Therefore, only one solution exists in this case: p=13, q=3.
Thus, the only solution which satisfies is p=13, q=3.
@Պոոռնապռագնյա Պռ
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Since p and q are primes, they both have 2 factors each: 1 and the number itself. This implies that p is equivalent to one of the expressions in the brackets.
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:
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Only one pair of primes exist. I used casework to solve this.
We have
p3−q7=p−q
p3−p=q7−q
p(p−1)(p+1)=q(q3−1)(q3+1)
p(p−1)(p+1)=q(q−1)(q+1)(q2+q+1)(q2−q+1)…(1)
I will first check all cases when p3<q7, that is, when p3−q7 is negative. Note that since p and q are primes, one of q, q−1, q+1, q2+q+1 or q2−q+1 is equal to p assuming q=2 (This issue will be addressed).
Since the difference is negative, we must have p<q. Combining this with what we know, we have p=q−1. (Note: p=q2−q+1 since the quadratic is greater than q for all positive q.) Since p and q are primes, we have p=2, q=3. Checking, we find that this is incorrect so there are no solutions in this case.
Now, we check when p=q. We get that p−1=q−1 and p+1=q+1, rendering q2+q+1=1 and q2−q+1=1, which gives p=q=0 which are not primes. There are no solutions in this case.
Finally, we check when p>q. We have the following values for p: p=q+1, p=q2−q+1, p=q2+q+1. We will now look at each expresion for p.
Case 1: p=q+1
We get that p=3, q=2. Checking, we find that this is incorrect. Therefore, there are no solutions in this case.
Case 2: p=q2−q+1
We have p=q2−q+1 so p−1=q2−q=q(q−1) which leaves p+1=(q+1)(q2+q+1) from (1). We now have
q2−q+1+1=(q+1)(q2+q+1)
q2−q+2=q3+2q2+2q+1
q3+q2+3q−1=0
Playing with this cubic, we find that it has no integral solutions for q (There are many different ways to check). Hence, there are no solutions in this case.
Case 3: p=q2+q+1
In this case, we have p=q2+q+1, so p−1=q2+q=q(q+1), which leaves p+1=(q−1)(q2−q+1) from (1). We now have the following:
q2+q+1+1=(q−1)(q2−q+1)
q2+q+2=q3−2q2+2q−1
q3−3q2+q−3=0
(q2+3)(q−3)=0
Clearly, only q=3 is the integral solution in this case, which implies p=13. Checking, we find that this is true. Therefore, only one solution exists in this case: p=13, q=3.
Thus, the only solution which satisfies is p=13, q=3.
Great problem, Surya.
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This is a great problem and solution!
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Same solution.
@Surya Prakash , is this a satisfactory solution?
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Respected Sharky, I did not understand the second sentence below equation 1 ( Note that since p & q are primes, ...) I request you to explain.
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Since p and q are primes, they both have 2 factors each: 1 and the number itself. This implies that p is equivalent to one of the expressions in the brackets.
No such pair of primes exist.
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No. There exists.
Check out p=13 and q=3.