<PRE><B>Question 316952</B>
Find three consecutive odd whole numbers such that the sum of their squares is a four-digit whole number whose digits are all the same
:
Three consecutive odd numbers x, (x+2), (x+4); resulting 4 number digit = y
:
x^2 + (x+2)^2 + (x+4)^2 = 1000y+100y+10y+y
:
x^2 + x^2+4x+4 + x^2+8x+16  = 1111y
:
x^2 + x^2 + x^2 + 4x + 8x + 4 + 16 = 1111y
:
3x^2 + 12x + 20 = 1111y
:
y = {{{(3x^2+12x+20)/1111}}}
:
Find the value for x which gives an integer value to y (the table of a TI83 was a great help!)
x=41, y=5
:
41^2 + 43^2 + 45^2 = 5555</PRE>