number theorysums-of-squares
"If n is a triangular number, show that each of the three consecutive integers, $8n^2, 8n^2+1, 8n^2+2$ can be written as a sum of two squares."
I have spend hours working on this problem and cannot seem to get anywhere with it. I was advised to start with $8n^2+1$ and work through it just using algebra to express this as a sum of two squares, but I am really struggling. Can anyone help?
THREE.
Multiply by 8. Add 3. It is a theorem of Gauss and Legendre that every $8n+3$ is the sum of three odd squares.
SEE, from the book The Sensual Quadratic Form by John Horton Conway, pages 138,139,140:
A list from a book by Dickson, giving sums $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 \leq a \leq b \leq c,$ and all the numbers that cannot be expressed by that one. So, $a=b=c=1,$ we find that the sum of three squares $x^2 + y^2 + z^2$ represents every positive integer not of the form $4^k \, (8n+7).$ In particular, all $8n + 3$ are represented.
One of the first identities you are likely to see in a discussion of complex numbers is the multiplicativity of the norm: $$ |zw| = |z||w|. $$ If you work that out in terms of real and imaginary parts of the constituents you see an interesting identity concerning sums of squares of real numbers. (That may be how you proved the multiplicativity of the norm.)
In this particular exercise the integers $2, 6, 8, 10, 13$ and so on happen to be just those integers that are the squares of norms of complex integers --- for example, $13 = |3+2i|^2 = 3^2 + 2^2$.
Best Answer
It is assumed in the books on this subject that all square $a^2$ is a sum of two squares $a^2+0^2$.
► Note first that for all $n$ one has $$8n^2=(2n)^2+(2n)^2\qquad (1)$$ Now all triangular number $n$ is of the form $$n=\frac{m(m+1)}{2}$$ Hence $$8n^2=\frac{8m^2(m+1)^2}{4}=2(m^2+m)^2$$ It follows the two identities
$$2(m^2+m)^2+1=(m^2-1)^2+(m^2+2m)^2\qquad (2)$$ $$2(m^2+m)^2+2= (m^2+m+1)^2+(m^2+m-1)^2\qquad (3)$$
With $(1),(2),(3)$ we have verified that the statement is true.