# Could square minus square be a square? [closed]

I wondered if a square minus a square could be a square ? When I put question into equation, I have

aa - bb = cc (when condition is a > b) *I don't know what is the correct notation of a mathematical condition


Q1: Is that right, that we always got a rectangle ?

Q2: Is there some easy way how to prove this ?

Q3: If answer to Q1 is true, is it also true when we remove the condition ?

Update

Q1 does not make sense, because we always get two rectangles, not one. But we can join these two rectangles to composed one. (which can have same square area as some other square)

• $25-16$ is a square. Apr 11, 2020 at 11:19
• $5^2-4^2=3^2$ or $13^2-12^2=5^2$, ecc. Apr 11, 2020 at 11:21
• There are many general cases. For example: for all $a$ and $b$, we have $$(a^2+b^2)^2-(a^2-b^2)^2=(2ab)^2$$ or for all $c$, we have $$(2c^2\pm 2c+1)^2-(2c^2\pm 2c)^2=(2c\pm 1)^2$$ (where the $\pm$ signs are not independent of each other). And it doesn't just stop there with the difference of two squares equalling a single square. Such general equations can involve abundances of square numbers! :) Apr 11, 2020 at 11:28
• I have no idea what Q1 means. What does it mean to "get a rectangle" here??? Apr 11, 2020 at 11:52
• @DavidC.Ullrich you are right, I updated my question. Apr 11, 2020 at 12:26

If $$a^2-b^2 = c^2$$, then $$a^2=b^2+c^2$$. The triples $$(a,b,c)$$ with this property are called Pythagorean triples. There are infinitely many of them.
• @Mr Pie But, the rectangle with square area 16 (where a = 2 and b = 8) has same square area as the square where a = 4. So the square area is the same, but one is rectangle and one is square, right ? Apr 11, 2020 at 11:36