|
|
| Line 1: |
Line 1: |
| | ===Problem=== | | ===Problem=== |
| | Find all pairs of primes <math>(p,q)</math> for which <math>p-q</math> and <math>pq-q</math> are both perfect squares. | | Find all pairs of primes <math>(p,q)</math> for which <math>p-q</math> and <math>pq-q</math> are both perfect squares. |
| − |
| |
| − | ==Solution 1==
| |
| − | We first consider the case where one of <math>p,q</math> is even. If <math>p</math> equals 2, <math>p-q=0</math> which gives us <math>pq-q=2</math> which doesn't satisfy the problem restraints. If <math>q</math>=2, we can set <math>p-2=x^2</math> and <math>2p-2=y^2</math> giving us <math>p=y^2-x^2=(y+x)(y-x)</math>. This forces <math>y-x=1</math> so <math>p=2x+1</math> and <math>p=x^2+2</math>. We then have <math>2x+1=x^2+2 \rightarrow x=1 \rightarrow (p,q)=(3,2)</math>.
| |
| − |
| |
| − |
| |
| − | Now assume that <math>p,q</math> are both odd primes. Set <math>p-q=x^2</math> and <math>pq-q=y^2</math> so <math>(pq-q)-(p-q)=y^2-x^2 \rightarrow p(q-1)=(y+x)(y-x)</math>. Since <math>y+x>y-x</math>, <math>p | (x+y)</math>. Note that <math>q-1</math> is an even integer and since <math>y+x</math> and <math>y-x</math> have the same parity, they both must be even. Therefore, <math>x+y=pk</math> for some even integer <math>k</math>. On the other hand, <math>p>p-q=x^2 \rightarrow p>x</math> and <math>p^2-p>pq-q=y^2 \rightarrow p>y</math>. Therefore, <math>2p>x+y</math> so <math>x+y=p</math>, giving us a contradiction.
| |
| − |
| |
| − | Therefore, the only solution to this problem is <math>(p,q)=(3,2)</math>.
| |
Revision as of 14:14, 3 August 2023
for which 
and 
are both perfect squares.
