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Another interesting number theory tidbit 
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#1
Jun2612, 04:44 PM

P: 63

Hello,
I was browsing a set of number theory problems, and I came across this one: "Prove that the equation a^{2}+b^{2}=c^{2}+3 has infinitely many solutions in integers." Now, I found out that c must be odd and a and b must be even. So, for some integer n, c=2n+1, so c^{2}+3=4n^{2}+4n+4=4[n^{2}+n+1]. If n is of the form k^{2}1, then the triple of integers{2n,2[itex]\sqrt{n+1}[/itex],2n+1]} satisfies the equation. Since there are infinitely such n, the equation holds for integers infinitely often. I thought this was cool. Mathguy 


#2
Jun2712, 02:52 AM

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Looks like the same approach could be used for many constants. So the question becomes, for what k does a^{2}+b^{2}=c^{2}+k have infinitely many solutions?



#3
Jun2712, 02:52 AM

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That is cool!!! Nice find!!
A nobrainer as followup question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out. 


#4
Jun2712, 03:22 AM

P: 295

Another interesting number theory tidbit
The case when k=0 has infinitely many solutions of which are all of the form [itex]a=d(p^2q^2)[/itex], [itex]b=2dpq[/itex], [itex]c=d(p^2+q^2)[/itex] for integer p,q and an arbitrary constant d. The case k=3 makes the right hand side the square of 2n+2 when c=2n+1, and hence the case k=0 implies the case k=3. Applying the case when k=0 that I specified above, I obtain that [itex]a=d(p^2q^2)[/itex], [itex]b=2dpq[/itex], [itex]c=d(p^2+q^2)1[/itex], which are, I believe, all of the solutions. However, note that if a particular selection of p and q yields c as even, then this will not hold. In particular, we need the above specified condition that [itex]n=k^21[/itex], so [itex]c=2k^21[/itex]. 


#5
Jun2712, 09:20 AM

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#6
Jun2712, 09:22 AM

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#7
Jun2712, 05:55 PM

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a = k + 2t + 1 (so a and k have opposite parity) b = (a^{2}  k  1)/2 c = b + 1 c^{2}  b^{2} = 2b+1 = a^{2}  k 


#8
Jun2712, 07:40 PM

P: 63




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