Let #D= a^2+b^2+c^2# where a and b are successive positive integers and #c=ab#.How will you show that #sqrtD# is an odd positive integer?

1 Answer
Jun 20, 2016

Answer:

#D = (a^2+a+1)^2# which is the square of an odd integer.

Explanation:

Given #a#, we have:

#b = a + 1#

#c = ab = a(a+1)#

So:

#D = a^2+(a+1)^2+(a(a+1))^2#

#=a^2+(a^2+2a+1)+a^2(a^2+2a+1)#

#=a^4+2a^3+3a^2+2a+1#

#=(a^2+a+1)^2#

If #a# is odd then so is #a^2# and hence #a^2+a+1# is odd.

If #a# is even then so is #a^2# and hence #a^2+a+1# is odd.