Find all primes such that \(p^3 - q^7 = p-q\).
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1 year, 11 months ago
Only one pair of primes exist. I used casework to solve this.
\[p(p-1)(p+1)=q(q-1)(q+1)(q^2+q+1)(q^2-q+1) \quad \ldots (1)\]
I will first check all cases when \(p^3 < q^7\), that is, when \(p^3-q^7\) is negative. Note that since \(p\) and \(q\) are primes, one of \(q\), \(q-1\), \(q+1\), \(q^2+q+1\) or \(q^2-q+1\) is equal to \(p\) assuming \(q \neq 2\) (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: \(p \neq q^2 - q + 1\) 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 \(q^2+q+1=1\) and \(q^2-q+1=1\), 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 \(p\): \(p=q+1\), \(p=q^2-q+1\), \(p=q^2+q+1\). 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: \(p=q^2-q+1\)
We have \(p=q^2-q+1\) so \(p-1=q^2-q=q(q-1)\) which leaves \(p+1=(q+1)(q^2+q+1)\) from \((1)\). We now have
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: \(p=q^2+q+1\)
In this case, we have \(p=q^2+q+1\), so \(p-1=q^2+q=q(q+1)\), which leaves \(p+1=(q-1)(q^2-q+1)\) from \((1)\). We now have the following:
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!
@Surya Prakash , is this a satisfactory solution?
Respected Sharky, I did not understand the second sentence below equation 1 ( Note that since p & q are primes, ...) I request you to explain.
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.
No. There exists.
Check out p=13 and q=3.