Suppose P and Q are both > 0. Then P >= 2Q+1 since the next perfect square after Q² is Q² + 2Q + 1. (We've assumed P>0 or else we'd also be able to consider squares less than Q²). Similarly, Q > 2P + 1. But the first inequality requires P>Q while the second Q>P so these can't both be true. Hence there are no solutions where P and Q are both > 0.
Now suppose P>0 but Q<0 (the same logic will apply to P<0 and Q>0). Then Q <= -2P+1 since the first available perfect square less than P² is P² - 2P + 1 and Q must be at least that negative. On the other side, P >= -2Q + 1. Writing both inequalities in terms of Q, we have Q <= -2P + 1 and Q >= (1-P)/2. These two expressions in P are equal when P= 1/3, and when P > 1/3, as when P is an integer >0, (1-P)/2 > -2P + 1. So we'd have the nonsensical result that -2P+1 >= Q >= (1-P)/2 AND (1-P)/2 > -2P+1. So there are no solutions when P>0 and Q<0. Swapping "P" and "Q" everywhere also demonstrates there are no solutions when P<0 and Q > 0.
The only case left is when P<0 and Q<0. Then the inequalities are: Q <= 2P+1 and P <= 2Q+1
Since P and Q are both negative, Q < 2P + 1 implies Q < P and likewise P >= Q. So the first inequality cannot hold. The only possibility is that Q = 2P+1, which imples P <= 4P + 3. Again, since P < 0 the inequality cannot hold, but the equality can -- P=4P+3 implies 3P+3 = 0 and P=-1, which in turn means Q = 2(-1) + 1 = -1.
And so the only solution possible is (-1,-1), which is consistent with the computer exploration earlier.
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Posted by Paul
on 2008-06-23 18:23:01 |
solution with proof
