# Very tentative proof of Beal's Conjecture?

I'm a high school student, so please point out my mistakes nicely and in layman's terms :) Thanks!

Ok. Beal's Conjecture: If

$$a^x+b^y=c^z$$ where $a$, $b$, $c$, $x$, $y$, $z$ are whole numbers; $x, y, z > 2$ ; and $a$, $b$, and $c$ are any natural number (now presumably WAY over than 2 based on counter-example searches); then a, b, and c must have at least one common factor.

Let's begin!

If x, y, and z are larger than 2, then the equation can be written thusly: $$a^2a^p+b^2b^q=c^2c^r$$ p, q, r being whole numbers.

Dividing by $c^r$ and taking the square root yields: $$a^2a^p/c^r+b^2b^q/c^r=c^2$$ $$a(a^p/c^r)^{1/2} +b(b^q/c^r)^{1/2}=c$$ Now, for a bit of logical analysis: If $r$ is not even, and thus $c^r$ is not a square, then $c$ would then be irrational because $c^r$ does not have a common factor with either $a^p$ or $b^q$, and thus, even if $a^p$ and $b^q$ were square, this would contradict the conjecture. The same reasoning applies to $a$ and $b$ themselves. Could $a^p, b^q, c^r$ be all not square? Nope. $$(m)^{1/2} + (n)^{1/2}= I$$ If $m$ and $n$ are not perfect squares and have no common factor, and I is an integer, this equation is impossible as $$m^2 + n^2 + 2(mn)^{1/2}=I^2$$ $2(mn)^{1/2}$ is obviously not rational, with our prerequisites. Therefore as the same reasoning occurs in the instance of $a^p$, $b^q$, and $c^r$ being non-squares, we can discard it, and we are left with the conclusion that they are all squares.

Meaning we can rewrite the original theorem from this: $$a^2a^p + b^2b^q=c^2c^r$$ into $$a^{2(e+1)} + b^{2(f+1)}=c^{2(g+1)}$$ where
$p=2e$
$q=2f$
$r=2g$ AND at the same time from the previous equation:$$a(a^p/c^r)^{1/2} +b(b^q/c^r)^{1/2}=c$$ into $$a(a^e/c^g) + b(b^f/c^g)=c$$ translating into $$a^{e+1} + b^{f+1}=c^{g+1}$$ Compare the transformations of the above two original equations. If we blend the results of both of them together that would mean:$$a^{2e+2} + b^{2f+2}=(a^{e+1} + b^{f+1})^2$$ Which is:$$2a^{e+1}b^{f+1}=a^{2e+2}-a^{2e+2}+b^{2f+2}-b^{2f+2}$$ $$2a^{e+1}b^{f+1}=0.$$ $$ab=0$$ Therefore if at least one of them must be zero, this contradicts the prerequisite that it must be a natural number. Proving Beal's million-dollar conjecture. I know it seems so easy, so it seems there must be some mistake.

• $\sqrt{x+y} \neq \sqrt{x}+\sqrt{y}!$ May 31, 2015 at 14:43
• It is good you are interested in such things. However, trying to prove well-known open problems is such a way is in my opinion quite a waste of time.
– quid
May 31, 2015 at 15:03
• An elaboration of @mt_ 's comment: After dividing both sides by $c^r$, you say "and taking the square root...." In that step you appear to be taking the square root of the individual terms of the left-hand-side, not the entire LHS. May 31, 2015 at 15:05
• And I respectully disagee with @quid's comment: While it may be a waste of time if your goal is to find a proof, it may nonetheless be very valuable as a learning experience to explore why it is so hard to prove. May 31, 2015 at 15:06
• @mweiss "be very valuable as a learning experience to explore why it is so hard to prove" Not even this in my firm opinion. (But let us better stop this discussion here.)
– quid
May 31, 2015 at 15:09

The mistake in the first step is pointed out by m_t on 5/31/15

$\sqrt{x+y}\ne\sqrt{x}+\sqrt{y}$

but looking at the end of the proof geometrically is also instructive.

Regardless of the mistake or the final exponents, the more general pattern is the equation for an arbitrary circle centered at the origin represented by

$a^2+b^2=c^2$

This circle always has four points where it intersects the $x$ or the $y$ axis. At these four points either $a$ or $b$ is zero. So, $ab=0$ at these four points. Now one can always draw a line of slope $1$ represented by

$a+b=c$

through one of those points. Rather than a contradiction, what is being described when you substitute the equation for the line into the equation for the circle is to see that they intersect.