Alright. So we want to solve the following equation for n:
n^2 + (n+1)^2 = m^2
where m is an integer. After a bit of manipulation we arrive at:
2n^2 + 2n + (1 + m^2) = 0
The solutions of this equation are:
n = (-2 +/- sqrt(4 - 8(1 + m^2))/4)
Or:
n = 1/2 (-1 +/- sqrt(2m^2 -...
I've seen Pell's equation also written as nx^2 - 1 = y^2, so maybe there are two forms, one with a plus sign, one with a minus sign. One fact about Pell's equation is that there are an infinite number of positive integer pair solutions (x,y) when n is not a perfect square integer (hopefully this...
Are you sure they're patternless? Just look at the odd n for now:
n=3: n^4+4 = 5*17
n=5: n^4+4 = 17*37
n=7: n^4+4 = 5*13*37 = 37*65
And I'll add n=9 for you:
n=9: n^4+4 = 5*13*101 = 65*101
See any pattern yet? That should lead you to the general proof.