Another interesting number theory tidbit

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$\sqrt{n+1}$,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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 Recognitions: Homework Help Science Advisor 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?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

Another interesting number theory tidbit

 Quote by micromass 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 $a=d(p^2-q^2)$, $b=2dpq$, $c=d(p^2+q^2)$ 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 $a=d(p^2-q^2)$, $b=2dpq$, $c=d(p^2+q^2)-1$, 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 $n=k^2-1$, so $c=2k^2-1$.

 Quote by haruspex 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.

 Quote by micromass 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)

Recognitions:
Homework Help
 Quote by Mathguy15 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

 Quote by haruspex 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!