The smallest positive integers which make $ p^2+10pq+q^2$ a perfect square are p=1, q=2. Is there an algorithm yielding larger values (if any)? Is there a way of determining how many solutions there are?
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1What have you tried? – Bumblebee Dec 01 '21 at 11:09
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By algebraic symmetry, every scaling of (1,2) is a solution. The same is true of any consecutive pair which gives a solution for $p^2+2kpq+q^2$. – Paul Stephenson Dec 02 '21 at 07:59
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(5,6) is also a solution in the original case. All I need in the general case is for $2(k+1)p(p+1)$ to be 8 X a triangular number. – Paul Stephenson Dec 02 '21 at 10:10
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How about editing that into your post, as well as reading the other post. – Bumblebee Dec 02 '21 at 10:52
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Sorry, the post quoted answers my question. My post withdrawn. – Paul Stephenson Dec 02 '21 at 11:45