How can I find all natural numbers a &b s.t. a^2 + a + 1 = b^2? Can anyone help me?

Answer 1

I assume natural numbers do not include 0. If 0 is allowed, then obviously a = 0 , b = 1 is a ttrivial solution. If 0 is not allowed, then I think there is no solution because:
a^2 + a + 1 = b^2
a^2+a = b^2-1
a^2 +a +1/4 = b^2 -3/4
(a+1/2)^2 = (4b^2 -3)/4
a+1/2 = +/- .5sqrt(4b^2-3)
a = -.5 +/- .5 sqrt(4b^2-3)
Thus 4b^2-3 must be an odd perfect square
Unfortunately there is no b meeting this requirement. I can't prove it, but i ran it on a spreadsheet for several thousand values and could not find a solution.

Answer 2

Here is an alternative solution. Convince yourself that for positive integral a the following is true:

a^2 < a^2 + a + 1 < (a+1)^2.

That means that a^2 + a + 1 lies between the squares of two consecutive integers. Therefore it cannot be a perfect square. If you include 0 as a natural number, then a^2 + a + 1 = 1 when a=0. It follows that b=1.

In sum, if you consider 0 to be a natural number then the only solution for (a,b) is (0,1). However if you exclude 0 from the set of natural numbers, then there is no solution in the natural numbers to the eqn above.

Answer 3

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)

Hope this hepled.

:)

Answer 4

The left hand side gives us a square of an imaginary number, but the right hand side gives us the square of a real number. This contrdiction indicates that there is no natural number to satisfy the equation.

Answer 5

(a+1/2)^2+3/4=b^2
(a+1/2)^2 - b^2 = -3/4
(a+1/2+b)(a+1/2-b) = -3/4

take a+1/2=c

(c+b)(c-b) = -3/4
(b+c)(b-c) = 3/4
b^2 - c^2 = 3/4

we know 1-1/4 = 3/4
b=1, c=1/2
but a+1/2=c
so a=0 and b = 1

also we can have b= -1 and c= -1/2
then a = -1 and b = -1