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?

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Answer 1 · 2016-06-28T21:45:09

See below

Making #a=n# and #b = n+1# and substituting in

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

which gives

#1 + 2 n + 3 n^2 + 2 n^3 + n^4#

but

#1 + 2 n + 3 n^2 + 2 n^3 + n^4 = (1 + n + n^2)^2#

which is the square of an odd integer