Add 3+7=10 & 5+9=14

10^2+11^2+12^2=13^2+14^2

10+11=21or14+13=27gives first and last numbers

So

21^2+22^2+23^2+24^2=25^2+26^2+27^2

Again 21+15=36 & 27+17=44

So

36^2+37^2+38^2+39^2+40^2=41^2+42^2+43^2+44^2

Next will be 36+19=55 and 44+21=65

No will be 55,56,57,58,59,60,61,62,63,64,65

And so on….

]]>Check this out about this equation: http://abstrusegoose.com/63

Loved it.

Alkis

]]>(-3) + 4 = 1² and (-3)² + 4² = 5²

9 + 40 = 7² and 9² + 40² = 41²

This has infinitely many solutions.

]]>There are solutions for longer strings of even consecutive integers.

It looks like there cannot be a solution in odd consecutive integers. ]]>