Show that any square number $k^2$ can be written as the sum of two squares and difference of two other squares

Question

While reading up on Pythagorean quadruples and Legendre's three square theorem, I encountered the following problem:

Since there are an infinite number of Pythagorean quadruples, it is true that the equation $a^2 + b^2 + c^2 = d^2$ has an infinite number of positive integer solutions. For example, $3^2 + 4^2 + 12^2 = 13^2$. In the same sense, we can show that $5^2$ can be written as the sum of two squares $3^2 + 4^2$ and the difference of two other squares $13^2 -12^2$. Using the results above, is it possible to show that any perfect square $k^2$ can be simultaneously written as the sum of two squares $a^2 + b^2$ and the difference of two other squares $d^2 - c^2?$

For a different version of the question, click here: Range of values of $k^2$ equal to the sum of two squares and the difference of two other squares.

Answer 1

While obviously, we have $k^2=k^2+0^2 = k^2-0^2$. We might want to exclude it to make things interesting and restrict each term to be non-zero.

In that case, it is not true.

Consider $2^2$, however, we can write it as the sum of two non-zero squares.

Non-zero squares that are smaller than it is $1$, hence it can't be written as the sum of two squares.

Similarly, $1^2=1$ cannot be the difference of two non-zero squares. Suppose $$1=c^2-b^2=(c-b)(c+b)$$

then we have $c-b=1$ and $c+b=1$, resulting in $c=1, b=0$.

Similarly, $2^2=c^2-b^2=(c-b)(c+b)$

Then we either have $(c-b, c+b)=(1,4)$ or $(c-b, c+b)=(2,2)$. The second case would lead to $b=0$. hence just consider the first case.

$$c-b=1$$ $$c+b=4$$

but adding them up would lead to a contradiction in parity.

Answer 2

"OP" needs solution for below mentioned simultaneous equation:

$a^2+b^2=w^2$

$c^2-d^2=w^2$

Take, $(a,b)=[(m^2-n^2),(2mn)]$

$(c,d)=[(p^2+q^2),(2pq)]$

Hence, $w^2=(m^2+n^2)^2=(p^2-q^2)^2$

So we impose the condition:

$m^2+n^2=p^2-q^2$

For, $(m,n,p,q)=(12,9,17,8)$ we get:

$(a,b,c,d,w)=(63,216,353,272)$

And,

$63^2+216^2=225^2$

$353^2-272^2=225^2$

Answer 3

We have:

$a^2+b^2=w^2$ -----(1)

$c^2-d^2=w^2$ ----(2)

taking:

$(a,b)=[(m^2-n^2),(2mn)]$ ----(3)

$(c,d)=[(p^2+q^2),(2pq)]$ ----(4)

$w^2=(m^2+n^2)^2=(p^2-q^2)^2$

$(m^2+n^2)=(p^2-q^2)$ ---------(5)

Parameterizing eqn (5) at,

$(m,n,p,q)=(2,1,3,2)$ we get:

$(m,n,p,q)=[(2t^2-4t+2),(t^2-4t+4),(3t^2-8t+6),(2t^2-6t+4)]$

Substituing above value's in (3) & (4) we get:

$a=(t^2-2)(3t^2-8t+6)$

$b=4(t^2-3t+2)^2$

$c=(13t^4-72t^3+152t^2-144t+52)$

$d=4(t^2-3t+2)(3t^2-8t+6)$

$w=(5t^4-24t^3+48t^2-48t+20)$

For, t=3 we get:

$(a,b,c,d,w)=(63,16,97,72,65)$

Hence simultaneous equation's (1) & (2) have

solution's above without any condition's on

the variables $(a,b,c,d,w)$.