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

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.

• Changing the nature of a question after it has been answered is bad form, as it makes the answerer appear to have misunderstood what you've asked, so I have rolled the question back to its original form. You should post your revised question separately. (It's free!) ... Also, please be clear in the question whether you are posing a conjecture, a problem of your own devising, etc. It's helpful for answerers to know how open-ended a question might be. – Blue May 3 '20 at 6:46
• Ahh I see sorry then I'll post another question with the edited version. Sorry! – Darrell Tan May 4 '20 at 12:06
• No harm done. :) ... You could/should update this question with a link to the new one, and have a link in the new question to this one. This will help move attention off of the old question, provide context for the new question, and avoid possible close-as-duplicate votes from people who don't notice the subtle differences between the two. Cheers! – Blue May 4 '20 at 12:10

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.

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

• Ah that makes sense I should've restricted the values to be nonzero positive integers and completely overlooked k=1 and k=2. Thanks! – Darrell Tan May 3 '20 at 6:33
• However, if I change the question to what are the set of values of k such that this is true, how would I tackle such a problem? – Darrell Tan May 3 '20 at 6:38
• I might start from studying Pythagoren triple but I'm not good with those. – Siong Thye Goh May 3 '20 at 8:26

"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,

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

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

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