Another interesting number theory tidbit


by Mathguy15
Tags: interesting, number, theory, tidbit
Mathguy15
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#1
Jun26-12, 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 a2+b2=c2+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 c2+3=4n2+4n+4=4[n2+n+1]. If n is of the form k2-1, 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
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haruspex
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#2
Jun27-12, 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 a2+b2=c2+k have infinitely many solutions?
micromass
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#3
Jun27-12, 02:52 AM
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That is cool!!! Nice find!!

A no-brainer as follow-up question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out.

Millennial
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#4
Jun27-12, 03:22 AM
P: 295

Another interesting number theory tidbit


Quote Quote by micromass View Post
That is cool!!! Nice find!!

A no-brainer as follow-up question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out.
That is an interesting question.

The case when k=0 has infinitely many solutions of which are all of the form [itex]a=d(p^2-q^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^2-q^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^2-1[/itex], so [itex]c=2k^2-1[/itex].
Mathguy15
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#5
Jun27-12, 09:20 AM
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Quote Quote by haruspex View Post
Looks like the same approach could be used for many constants. So the question becomes, for what k does a2+b2=c2+k have infinitely many solutions?
Hm, I realize my approach works for k congruent to 3(mod4), but beyond that, I don't know.
Mathguy15
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#6
Jun27-12, 09:22 AM
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Quote Quote by micromass View Post
That is cool!!! Nice find!!

A no-brainer as follow-up question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out.
Haha, Thanks! I'm inclined to say that these are all the solutions, but I do not know. By the way, I could write the solutions in terms of k rather than n to make it neater. (i.e., if k is an integer, then, {2k2-2,2k,2k2-1} is an integer solution to the equation)
haruspex
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#7
Jun27-12, 05:55 PM
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Quote Quote by Mathguy15 View Post
Hm, I realize my approach works for k congruent to 3(mod4), but beyond that, I don't know.
Choose any t > 0.
a = k + 2t + 1 (so a and k have opposite parity)
b = (a2 - k - 1)/2
c = b + 1
c2 - b2 = 2b+1 = a2 - k
Mathguy15
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#8
Jun27-12, 07:40 PM
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Quote Quote by haruspex View Post
Choose any t > 0.
a = k + 2t + 1 (so a and k have opposite parity)
b = (a2 - k - 1)/2
c = b + 1
c2 - b2 = 2b+1 = a2 - k
Brilliant!


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