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# Olympiad Proof Problem - Day 4

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

Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
1 year, 9 months ago

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Only 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. · 1 year, 9 months ago

This is a great problem and solution! · 1 year, 9 months ago

Same solution. · 7 months, 3 weeks ago

@Surya Prakash , is this a satisfactory solution? · 1 year, 9 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. · 1 year, 9 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. · 1 year, 9 months ago

No such pair of primes exist. · 1 year, 9 months ago

No. There exists. · 1 year, 9 months ago