Is it true that, for any Pythagorean triple $4ab > c^2$?

Is it true that, for any Pythagorean triple $$4ab > c^2$$?

So this came up in a proof I was working on and it seems experimentally correct from what I've tried and I would imagine the proof is similar to proving,

$$ab < \frac{c^2}{2}$$

The idea I have for this approach is,

$$4ab > c^2$$ $$4ab > a^2 + b^2$$ (Then maybe something with the triangle inequality?)

So is this statement true (it seems to be), and how can I prove it?

A counter-example would also be acceptable.

• I don't think it will be true if the triangle is long and thin. Dec 3 '19 at 19:34
• I'll do some more testing with that in mind thanks Dec 3 '19 at 19:35
• A counter example would also be acceptable Dec 3 '19 at 19:36

$$15^2 + 112^2 = 113^2$$, but $$4 \cdot 15 \cdot 112 \le 113^2$$.

• Thanks! I really didn't want it to be true because it would have resolved another question I was working on. None the less greatly appreciative. Dec 3 '19 at 19:41

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

$$c^2 < 4ab$$ is the same thing as claiming $$a^2 + b^2 < 4ab$$

This is true if and only if $$a^2 - 2ab + b^2 < 2ab$$ which is to say

$$(a-b)^2 < 2ab$$

There shouldn't be any reason that should be true in general. If let $$m =$$ average/midpoint of $$a,b$$, that is, $$m = \frac {a+b}e$$ and $$e =$$ the radius of the interval $$a$$ to $$b$$, that is $$e = |m-a| =|b-m| = \frac{|a-b|}2$$ then we have

$$(a-b)^2 < 2ab \implies$$

$$4e^2 < 2(m-e)(m+e) = 2(m^2 - e^2)\implies$$

$$3e^2 < m^2$$

Which utterly need not be true!

Of course finding a pythagorean triplet where $$a= m\pm e$$ and $$b=m\mp e$$ are integers so that $$a^2 + b^2 = 2(m^2 + e^2)$$ is not merely and integer but a perfect square may put a wrinkle in finding a counterexample.

But not an insurmountable wrinkle.

We need, for a counter-example, $$3e^2 \ge m^2$$ which just require $$e$$ be significantly large, which is utterly irrelevent (it would seem) to $$2(m^2 + e^2)$$ being a perfect square.

... But to find a counter example:

.....

Back to square 1: We can find a Pythogorian triple $$a^2 + b^2 = c^2$$ by letting $$b$$ be any odd integer and if $$b^2 = 2a+1$$ so $$c^2 = (a+1)^2$$ ... retrosolving by letting $$a = \frac {b^2-1}2$$.

So we want a counter example of $$c^2 = a^2 + b^2 = a^2 + 2a + 1 = (a+1)^2 \ge 4ab$$ or substitutint $$a = \frac {b^2-1}2$$, we want a case where

$$(\frac {b^2+1}2)^2 \ge 4\frac{b^2 -1}2b$$.

Surely we can find a counter example.

$$(b^2 + 1)^2 \ge 8(b^2-1)b$$

$$b^4 + 2b^2 + 1 \ge 8b^3 -8b$$

$$b^4 -8b^3 +2b^2 - 8b + 1 \ge 0$$

yeah that can be solved...

(Obviously as $$b\to \infty$$ the $$b^4-8b^3 + 2b^2 - 8b + 1\to \infty$$ so there is some $$K$$ where $$b > K$$ will always have $$b^4 -8b^3 +2b^2 - 8b + 1\ge 0$$.)

Taking a sledgehammer and letting $$b > 8$$, say, $$b=9$$ and $$a = \frac {b^2 -1}2=40$$ so $$40^2 + 9^2 = 41^2> 4*9*40=36*40$$

Yeah... it's not true.

(Actually our sledgehammer was fairly precise. $$b^4 -8b^3 +2b^2 - 8b + 1$$ has only two real solutions; one between $$0$$ and $$1$$ (far closer to $$0$$ than to $$1$$) and the other between $$7$$ and $$8$$. So of primmitive triplets where $$b$$ is odd and $$b^2 = 2a+1$$. $$a = 40, b=9, c=41$$ is the smallest primitive triplet (and thus smallest of all triplets) where $$c^2 > 4ab$$.)

( so $$c^2 < 4ab \iff \frac {c}{\gcd(a,b,c)} < 41$$)

• Thanks! Also I wish I could give +2 for the sledgehammer as it makes a way to construct a counterexample! Certainly a useful tool Dec 3 '19 at 20:30
• Actually $a=40$ and $b = 9$ is the smallest primitive counterexample but all with $8< b< a$ will fail. Dec 3 '19 at 20:32