Day 13: A Prime Time for a Number Theory Problem

It is possible to bash this problem but here is another way:

Note that $(p-q)(p+q)=p^2-q^2$ . Now apart from $3$ , all primes squared are $1\, \mathrm{mod}\, 3$ , so if both primes are not $3$ , then the LHS must be divisible by $3$ , and so there cannot be a solution as $208$ is not divisible by $3$ . Also if $p=q=3$ it is clear that the equation is not satisfied.

Therefore one of $p$ and $q$ must be $3$ . If $p=3$ then $q=2$ , as otherwise the LHS is negative. This does not work. Therefore $q=3$ . Checking shows that $p=5$ works, and it is clear that the LHS will only increase as $p$ increases.

And so the only solution is $(5,3)$ giving the answer $\boxed{15}$ as required.

208=2^4×13 (p-q)(p+q)(2p+q)=208 let's assume that p and q are equal to 2. that leaves us with (p-q)=2k,(p+q)=2n,(2p+q)=2z+1 since there's only one option for a non even Factor for 208 we can say that 2p+q=13 and since p-q<p+q there is no other option than setting p-q=2 p+q=8. after choosing 2 out of the 3 eqaution and solving we see that p=5 q=3 setisfy all the conditions therefore q×p=15