A Diophantine Equation: “How can I find all natural numbers a &b such that a^2 + a + 1 = b^2?”

Answer

We assume “natural numbers here includes the number zero” (an old convention – may not be valid today!)

If not, there are no solutions!

We proceed as follows:

a^2 + a + 1 = b^2

Rewrite as,
a^2 + 2 a + 1 = b^2 + a
( a + 1 ) ^2 = b^2 + a
( a + 1 ) ^2 – b^2 = a
( a + 1 +b ) ( a + 1 – b ) = a

Now note that in the LHS ( Left Hand Side of the equation),
( a + 1 +b ) > a for all a > 0. In other words the LHS > RHS
( Right Hand side ) of the equation above for all a > 0. Hence a must be zero ( natural numbers are either 0 or the positive integers… this is the most widely accepted definition by convention ) !

a = 0. implies that b^2 = 1 yielding b =1 since b>0.

Hence the solution is : (a,b) = (0,1)