Another interesting number theory tidbit

[Mathguy15]

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

 

[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?

 

[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.

 

[Millennial]

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 $c=2k^2-1$.

 

[Mathguy15]

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]

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]

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]

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!