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Sorry to ask two questions but I would ask: What conditions will these two fractions$$\frac {p^2+t^2} {p-t} $$ and $$\frac {pt(p+t)} {p-t}$$ Both be integers. I found that not all the time the fraction can both be function. Obviously $t=p-1$ will work, but what if $t=p-n$? Note that both $p $ and $t$ are positive integers on both fractions

What have I tried: Assuming $t=p-n$ getting $\frac {2p^2-2pn+n^2} n$ and $\frac {a(a-n)(2a-n)} n$. Or assuming $t=pk$ and getting $\frac {p(k^2+1)} {1-k}$ and $\frac {p^2k(1+k)} {1-k}$. How can I go further than that?

xxxx036
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I called them $x,y$ and the ratio $z.$ There are two infinite families that make up all coprime integer solutions to $$ x^2 + y^2 + yz - zx = 0 $$ The first family is $$ x=u^2 + uv \, , \; \; y = -u^2 + uv \, , \; \; z = u^2 + v^2 $$ For this family $$ \frac{xy(x+y)}{x-y} $$ is always an integer. There is, in particular, no need to investigate common factors anywhere.

The second family is $$ x=2u^2 + 3uv \, , \; \; y = u^2 + uv \, , \; \; z = 5u^2 +4uv v^2 $$ Let me look at that again, I don't appear to have written it out

For this family $$ \frac{xy(x+y)}{x-y} = \frac{u^2 (u+v)(2u+3v)(3u+4v)}{u+2v} $$ and this may not be an integer. It ought to be possible, and quick, to check the individual conditions, (pretending the the denominator is prime) and working out the possible patterns for $u,v.$ Note that we can take coprime $u,v$ and just multiply both by $u+2v,$ since the new denominator is $(u+2v)^2$ but the numerator has been multiplied by $(u+2v)^5$

BACKGROUND:

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Will Jagy
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