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Suppose p and q are both prime numbers where p>q. Show that p-q and p+q cannot both be perfect squares.


  1. Assume that q≥3. If p+q and p−q are both squares, then p−q and p+q are both even squares, which means both are divisible by 4, which means then that p+q−(p−q)=2q must be divisible by 4. But this cannot be i.e., 2q=(p+q)−(p−q) cannot be divisible by 4, as q is an odd number as q is a prime at least 3.
    [Note that I only used that p and q are odd.]
    As for q=2, note that with the exception of the pairs 1,4, the absolute value of the difference between every two integral squares is at least 5. So with q=2, note that p+2 and p−2 cannot both be squares as they differ by only 4<5.


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