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Suppose we two equations $$x^2+y^2=r^2$$ and $$(x-a)^2+(y-b)^2=2g^2$$

Where x,y and r are integer variables greater than 0. a,b and g are integer constants greater than 0.

I conjecture that for any given selection of a,b and g, the number of integer solutions of the form $(x,y,r)$ is less than or equal to 3. How would I go about proving or disproving this? Any tips on tackling this problem. Thanks.

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  • $\begingroup$ Hi lulu. The radius r is a variable. So the first equation is actually giving an infinite set of circles that intersect with the circle in the second equation. $\endgroup$ – Ameet Sharma Nov 9 '15 at 19:12
  • $\begingroup$ Thanks! I worked that out and then removed my comment. $\endgroup$ – lulu Nov 9 '15 at 19:13
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How to get circles with more than $3$ such points...

First, consider "generating" circle of the form $$ x_0^2+y_0^2=2g^2, $$ that has many integer points. One of such rather small and "reach on integer points" circles is the circle with $g=1105=5\times 13\times 17$. It has $108$ integer points (for example).

Second, shift this circle at (any random) integer vector $(a,b)$ and check how many its integer points match condition $$x^2+y^2=r^2,$$ where $r$ is integer.

Then for considered circle and for $(a,b)=(10, 1855)$ (for example) one will get system $$ \left\{ \begin{array}{c} (x-10)^2+(y-1855)^2=2\cdot 1105^2,\\ x^2+y^2=r^2, \end{array} \right. $$ which has $4$ such $(x,y,r)$-solutions:
$(x,y,r)=(231,308,385)$,
$(x,y,r)=(377,336,505)$,
$(x,y,r)=(1557,2076,2595)$,
$(x,y,r)=(605,3300,3355)$.

And I believe that there are examples with more than $4$ points.


Ooops... Consider the same $g=1105$, and $(a,b)=(73,1561)$.
This construction gives us $7$ solutions:
$(x,y,r)=(440,42,442)$,
$(x,y,r)=(882,224,910)$,
$(x,y,r)=(1410,752,1598)$,
$(x,y,r)=(1592,1194,1990)$,
$(x,y,r)=(1634,1488,2210)$,
$(x,y,r)=(1224,2618,2890)$,
$(x,y,r)=(1130,2712,2938)$.


Update:

for $g=32045=5\times 13\times 17\times 29$ (circle with $324$ integer points)
one can find $(a,b)=(823,45311)$, which give us $15$ solutions:
$(x,y,r)=(12720, 1582, 12818)$,
$(x,y,r)=(25382, 7224, 26390)$,
$(x,y,r)=(32144, 12558, 34510)$,
$(x,y,r)=(33576, 13990, 36374)$,
$(x,y,r)=(38910, 20752, 44098)$,
$(x,y,r)=(44552, 33414, 55690)$,
$(x,y,r)=(46134, 44488, 64090)$,
$(x,y,r)=(46040, 48342, 66758)$,
$(x,y,r)=(45080, 55062, 71162)$,
$(x,y,r)=(40062, 67984, 78910)$,
$(x,y,r)=(35064, 74998, 82790)$,
$(x,y,r)=(30510, 79552, 85202)$,
$(x,y,r)=(23496, 84550, 87754)$,
$(x,y,r)=(10574, 89568, 90190)$,
$(x,y,r)=( 3854, 90528, 90610)$.

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  • $\begingroup$ Thanks! How did you find a 'g' with many integer points? $\endgroup$ – Ameet Sharma Nov 10 '15 at 1:23
  • $\begingroup$ One can find such small $g$ via brute force (loops on $g$ and checking appropriate $x,y$). But this is closely related to Gaussian Integers. That's why decomposition of $1105$ has all prime numbers of the form $p=4k+1$ ($5=4\times 1+1$, $13=4\times 3+1$, $17=4\times 4+1$). $\endgroup$ – Oleg567 Nov 10 '15 at 1:27
  • $\begingroup$ Can you find such $(a,b)$ for this $g=1105$, that give $8$ integer solutions? I am sure that such $(a,b)$ exist ;) . $\endgroup$ – Oleg567 Nov 10 '15 at 1:30
  • $\begingroup$ @Ameet, I updated my answer with $(g,a,b)$ that generates $15$ solutions $(x,y,r)$. $\endgroup$ – Oleg567 Nov 10 '15 at 20:39
  • $\begingroup$ Thanks Oleg567. Really appreciate it! Looks like my conjecture was way off. :) I will write my own program to generate these. $\endgroup$ – Ameet Sharma Nov 10 '15 at 20:43

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