# About three conjectures concerning the Pythagorean Theorem.

I have two main conjectures relating to the Pythagorean Theorem that I desperately want to find out if they are true or not. Could somebody please help me?

• If $\ a^2 + b^2 = c^2 \$ for which $c$ is a prime number, then for some $n \in \mathbb{N}, \ c = 4n + 1$.
• $\forall \{x, y, z\} \subset \mathbb{N}, \ \big(x^2 + (x + 1)^2\big)^2 = y^2 + z^2 \$ for which $x \neq y \neq z$.

Tested for $c, x \leqslant 10^6$.

I did also have another conjecture that if $\ a^2 + b^2 = c^2 \$ for which $a$ is odd (and greater than $1$), $b$ is even, and $a < b$ then there existed a solution such that $c = b + 1$, but I am $100$% certain this has been discovered before (assuming so, since it was not very difficult to notice). However I am not sure if the other two conjectures have been discovered before, although if it had to be one of them, I would pick the second one. So could somebody also please clarify that for me?

Thank you in advance.

• Not sure what the second one means (as stated it is clearly false). In general we have $(a^2+b^2)\times (c^2+d^2)=(ac+bd)^2+(ad-bc)^2$. Is that what you are getting at? – lulu Oct 17 '17 at 14:41
• I think they mean that any $(x^2 + (x+1)^2)^2$ can be written as the sum of two squares, which I imagine will come directly from Fermat's two-squares theorem – TheMathsGeek Oct 17 '17 at 14:43
• As a general remark, you should take greater care in stating your assumptions. For the first, the statement is clearly false unless you require both $a,b>0$. For instance, $0^2+7^2=7^2$. I assume you meant to exclude that but you didn't do so explicitly. For the second, your statement starts with "for all $x,y,z\in \mathbb N$". Well, then I should be able to take $x=1,y=1,z=1$ but $[1^2+2^2]^2=25\neq 1^2+2^2$. Again I'm sure you meant something else but it is important to state things clearly. – lulu Oct 17 '17 at 14:47
• @TheMathsGeek Oh, I think the same thing and the formula I provided shows that the product of any two sums of two squares is again the sum of two squares, which is a stronger claim. – lulu Oct 17 '17 at 14:48
• @lulu For sure, plus yours is definitely an easier answer to understand! – TheMathsGeek Oct 17 '17 at 14:49

Suppose that is $c=4k+3$. Then $c|a^2+b^2$ and since it is prime $c|a$ and $c|b$. Thus $a=mc$ and $b=nc$. But then we get $m^2+n^2=1$ which is impossible if $mn\ne 0$.

The accepted answer covers the first conjecture.

The second and third conjectures are in fact related. To solve,

$$\big(x^2 + (x + 1)^2\big)^2 = y^2+z^2$$

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

and turns out to have the solution,

$$\big(x^2 + (x + 1)^2\big)^2 = (2x + 1)^2+(\color{brown}{2x^2 + 2x})^2 =(\color{brown}{2x^2 + 2x}+1)^2$$

• Back when I asked this question, I didn’t know that. But I figured the rest of the other conjectures independently. Go here $\longrightarrow$ math.stackexchange.com/questions/2620072/… to see. Nice answer, though :) (I will upvote in at least $7$ hours) – Mr Pie Feb 14 '18 at 16:47