- #1
Mathguy15
- 68
- 0
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
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