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

\[q^2-q+1+1=(q+1)(q^2+q+1)\]

\[q^2-q+2=q^3+2q^2+2q+1\]

\[q^3+q^2+3q-1=0\]

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: \(p=q^2+q+1\)

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:

\[q^2+q+1+1=(q-1)(q^2-q+1)\]

\[q^2+q+2=q^3-2q^2+2q-1\]

\[q^3-3q^2+q-3=0\]

\[(q^2+3)(q-3)=0\]

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.
–
Sharky Kesa
·
1 year, 5 months ago

@Sharky Kesa
–
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
·
1 year, 5 months ago

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@Պոոռնապռագնյա Պռ
–
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.
–
Sharky Kesa
·
1 year, 5 months ago

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No such pair of primes exist.
–
Saakshi Singh
·
1 year, 5 months ago

## Comments

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

We have

\[p^3-q^7=p-q\]

\[p^3-p=q^7-q\]

\[p(p-1)(p+1)=q(q^3-1)(q^3+1)\]

\[p(p-1)(p+1)=q(q-1)(q+1)(q^2+q+1)(q^2-q+1) \quad \ldots (1)\]

I will first check all cases when \(p^3 < q^7\), 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

\[q^2-q+1+1=(q+1)(q^2+q+1)\]

\[q^2-q+2=q^3+2q^2+2q+1\]

\[q^3+q^2+3q-1=0\]

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:\(p=q^2+q+1\)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:

\[q^2+q+1+1=(q-1)(q^2-q+1)\]

\[q^2+q+2=q^3-2q^2+2q-1\]

\[q^3-3q^2+q-3=0\]

\[(q^2+3)(q-3)=0\]

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. – Sharky Kesa · 1 year, 5 months ago

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– Sualeh Asif · 1 year, 5 months ago

This is a great problem and solution!Log in to reply

– Abhi Kumbale · 3 months, 3 weeks ago

Same solution.Log in to reply

@Surya Prakash , is this a satisfactory solution? – Sharky Kesa · 1 year, 5 months ago

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– Պոոռնապռագնյա Պռ · 1 year, 5 months ago

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

– Sharky Kesa · 1 year, 5 months ago

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.Log in to reply

No such pair of primes exist. – Saakshi Singh · 1 year, 5 months ago

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– Surya Prakash · 1 year, 5 months ago

No. There exists.Log in to reply

– Sharky Kesa · 1 year, 5 months ago

Check out p=13 and q=3.Log in to reply