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| | ===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. |
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| − | ==Solution==
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| − | Let <math>p-q = a^2</math>, <math>pq - q = b^2</math>, where <math>a, b</math> are positive integers. <math>b^2 - a^2 = pq - q - (p-q) = pq -p</math>. So,
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| − | <cmath> b^2 - a^2 = p(q-1) \tag{1}</cmath>
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| − | <math>\bullet</math> For <math>q=2</math>, <math>p = b^2 - a^2 = (b-a)(b+a)</math>. Then <math>b-a=1</math> and <math>b+a=p</math>. <math>a=\dfrac{p-1}{2}</math> and <math>p-q = a^2</math>. Thus, <math>p - 2 = \left( \dfrac{p-1}{2} \right)^2 \implies p^2 - 6p + 9 = 0</math> and we find <math>p=3</math>. Hence <math>(p,q) = (3,2)</math>.
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| − | <math>\bullet</math> For <math>q=4k+3</math>, (<math>k\geq 0</math> integer), by <math>(1)</math>, <math>p(4k+2) = b^2 - a^2</math>. Let's examine in <math>\mod 4</math>, <math>b^2 - a^2 \equiv 2 \pmod{4}</math>. But we know that <math>b^2 - a^2 \equiv 0, 1 \text{ or } 3 \pmod{4}</math>. This is a contradiction and no solution for <math>q = 4k + 3</math>.
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| − | <math>\bullet</math> For <math>q=4k+1</math>, (<math>k > 0</math> integer), by <math>(1)</math>, <math>p(4k) = b^2 - a^2</math>. Let <math>k=m\cdot n</math>, where <math>m\geq n \geq 1</math> and <math>m, n</math> are integers. Since <math>p>q</math>, we see <math>p>4k</math>. Thus, by <math>(1)</math>, <math> (b-a)(b+a) = 4p\cdot m \cdot n</math>. <math>b-a</math> and <math>b+a</math> are same parity and <math>4p\cdot m \cdot n</math> is even integer. So, <math>b-a</math> and <math>b+a</math> are both even integers. Therefore,
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| − | <math>
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| − | \left\{ \begin{array}{rcr}
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| − | b+a = & 2pn \\
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| − | b-a = & 2m
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| − | \end{array} \right.
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| − | </math>
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| − | or
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| − | <math>
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| − | \left\{ \begin{array}{rcr}
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| − | b+a = & 2pm \\
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| − | b-a = & 2n
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| − | \end{array} \right.
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| − | </math>
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| − | Therefore, <math>a=pn - m</math> or <math>a = pm - n</math>. For each case, <math>p-q = p - 4mn - 1 < a</math>. But <math>p-q = a^2</math>, this gives a contradiction. No solution for <math>q = 4k + 1</math>.
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| − | We conclude that the only solution is <math>(p,q) = (3,2)</math>.
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| − | (Lokman GÖKÇE)
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for which 
and 
are both perfect squares.
