4
$\begingroup$

Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?

Could you show me the proof?

Improve this question
$\endgroup$
  • $\begingroup$ It takes a while. The result is due to Fermat. and is usually proved using "infinite descent." You can find a moderately detailed sketch here. $\endgroup$ – André Nicolas Oct 17 '13 at 17:19
  • $\begingroup$ So the Fermat's result has a name? $\endgroup$ – jachilles Oct 17 '13 at 18:06
  • $\begingroup$ Not really. It is closely connected to the case $n=4$ of Fermat's Last Theorem. Slight generalization. $\endgroup$ – André Nicolas Oct 17 '13 at 18:08
  • $\begingroup$ For a generalization, the congruent number problem seeks to solve $a^2+nb^2 = c^2$ and $a^2-nb^2=d^2$. Your question is just the case $n=1$. The list of solvable $n$ is $n=5, 6, 7, 13, 14,\dots$. $\endgroup$ – Tito Piezas III Oct 18 '13 at 2:14
  • $\begingroup$ Wow Thank you @TitoPiezasIII $\endgroup$ – jachilles Oct 18 '13 at 5:32
3
$\begingroup$

$${ \left( \frac { a }{ b } \right) }^{ 2 }+1={ \left( \frac { c }{ b } \right) }^{ 2 }\\ \frac { c }{ b } +\frac { a }{ b } =k\\ \frac { c }{ b } -\frac { a }{ b } =\frac { 1 }{ k } \\ \frac { c }{ b } =\frac { { k }^{ 2 }+1 }{ 2k } \\ \frac { a }{ b } =\frac { { k }^{ 2 }-1 }{ 2k } \\ $$Say:$$a=({ k }^{ 2 }-1)n\\ b=2kn\\ { a }^{ 2 }-{ b }^{ 2 }={ n }^{ 2 }({ k }^{ 4 }-6{ k }^{ 2 }+1)=n^2(k^2-3)^2-8=d^2$$ Is the result a perfect square?

Improve this answer
$\endgroup$
  • $\begingroup$ Hey.. I don't quite get where your second eqn has come from.. $\endgroup$ – jachilles Oct 17 '13 at 18:31
  • $\begingroup$ Could you please make it a bit more detailed and clarify what you are doing on the lines of the proof? $\endgroup$ – jachilles Oct 17 '13 at 21:19
  • $\begingroup$ @julypraise First I am not sure if the result is correct but it seems to me it may be. I have divided both sides of the equation with $b$. $(c/b)^2-(a/b)^2=1$, therefore if $c/b+a/b=k$ then $c/b-a/b=1/k$. Afterwards I have expressed $c,b,a$ in terms of $k$. $\endgroup$ – newzad Oct 17 '13 at 21:26
  • $\begingroup$ Thanks for your response. So $k$ is an integer? $\endgroup$ – jachilles Oct 17 '13 at 21:28
  • $\begingroup$ @julypraise no, it is a rational number. $\endgroup$ – newzad Oct 17 '13 at 21:29
2
$\begingroup$

Dividing by $(a,b)$, we can assume that $(a,b)=1$.

Multiplying the two equations yields $$ a^4-b^4=c^2d^2\tag{1} $$ whereby $$ b^4+(cd)^2=a^4\tag{2} $$

This answer characterizes all primitive Pythagorean Triples; that is, there must be a pair of positive integers $(m,n)$ so that $(m,n)=1$ and $m+n$ is odd so that $$ x=m^2-n^2,\quad y=2mn,\quad z=m^2+n^2\tag{3} $$ Suppose that $(x,y,z)$ is the Pythagorean triple with the smallest hypotenuse so that at least two sides are a square or twice a square.

Suppose that $z$ and $x$ are squares (since they are odd, they can't be twice a square). Then $(\sqrt{xz},n^2,m^2)$ has a smaller hypotenuse.

Suppose that $z$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(m,n,\sqrt{z})$ has a smaller hypotenuse (since $x,y\ge1$, $z\gt1$).

Suppose that $x$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(\sqrt{x},n,m)$ has a smaller hypotenuse.

Thus, there is no primitive Pythagorean Triple with at least two sides a square or twice a square.

If $(2)$ were true, there would be a primitive Pythagorean Triple with two sides a square.

Improve this answer
$\endgroup$
1
$\begingroup$

I'll borrow newzad's caclulations and we'll prove that that last equation can't be a square number. We have:

$$a^2 - b^2 = n^2(k^4 - 6k^2 + 1)$$

Because $n^2$ is a square of some natural number $n$ we need to prove that $k^4 - 6k^2 + 1$ is a square number in order RHS to be also a square. First make a substitution $k^2 = t$ to reduce it to quadratic equation. And because we want it to be a square, set it to $m^2$. So we have:

$$k^4 - 6k^2 + 1 = t^2 - 6t + 1 = m^2$$

$$t^2 - 6t + (1 - m^2) = 0$$

Now solve this quadratic equation with respect to $t$. Because all coefficients are integers it'll have a integer solution, if the discriminant is a square number so we have:

$$D = b^2 - 4ac = (-6)^2 - 4(1-m^2) = 36 - 4 + 4m^2 = 2^2(9 - 1 + m^2) = 2^2(m^2 + 8)$$

Again we have that the discriminant will be a square number if $m^2 + 8$ is a square so we have:

$$m^2 + 8 = l^2$$ $$m^2 - l^2 = -8$$ $$(m-l)(m+l) = -8$$

By checking for factors of $-8$ we get that $m=\pm1$ and $l=\pm3$ is the only integer solutions.

So we can subtitute back in the initial equation and we have:

$$t^2 - 6t + 1-1 = 0$$ $$t^2 - 6t = 0$$ $$t(t-6) = 0$$ $$t_1 = 0 \quad \quad \quad t_2 = 6$$

This implies that:

$$k_{1/2} = 0 \quad \quad \quad k_{3/4} = \pm\sqrt{6}$$

But none of this is possible, because in newzad's caculation we have:

$$\frac{a}{b} + \frac{c}{b} = k$$

Because $a,b,c \in \mathbb{N}$ it's impossible for $k=0$ and also because $a,b,c$ are positive integers we have that $\frac{a}{b}$ and $\frac{c}{b}$ are rational numbers, but we know that a sum of two rational numbers, will never be an irrational numbers, so it impossible $k = \pm \sqrt{6}$

Because there aren't any more possibilities for $k$ we have that it's impossible for $a,b,c,d \in \mathbb{N}$ the following to hold:

$$a^2 + b^2 = c^2 \quad \quad \quad a^2 - b^2 = d^2$$

Q.E.D.

Improve this answer
$\endgroup$
  • $\begingroup$ You consider $n$ as a natural number. I think it is not necessarily a natural number, rather it is a rational number but maybe I am wrong. $\endgroup$ – newzad Oct 17 '13 at 22:54
  • $\begingroup$ I think you're right, I'll edit my notation. $t^2 - 6t + 1$ need to be a square number of rational number. So we need to check all rational solutions for $(m-l)(m+l) = -8$ a rational numbers, rather than integers. I'll give it a shot. $\endgroup$ – Stefan4024 Oct 17 '13 at 22:58
  • $\begingroup$ I searched over the internet and I found a generator for rational solutions of this type of equation here and there are infinite amount of them, I'm starting to believe that this method is doomed to fail. $\endgroup$ – Stefan4024 Oct 17 '13 at 23:29
0
$\begingroup$

Summing and subtracting, you get $$\begin{cases} 2a^2=c^2+d^2\\2b^2=c^2-d^2\end{cases}$$ Thus $c,d$ have the same parity. If they are both even, we obtain $2a^2=4c'^2+4d'^2$ and $2b^2=4c'^2-4d'^2$, so $$\begin{cases} a^2=2(c'^2+d'^2)\\b^2=2(c'^2-d'^2)\end{cases}$$

Since $2\mid a^2,b^2$, we must have $a^2=4a'^2$, $b^2=4b'^2$, so we get that $$\begin{cases} 2a'^2=c'^2+d'^2\\2b'^2=c'^2-d'^2\end{cases}$$

And now you're back to the same, but with smaller numbers. And thus we descend.

It remains the case that both are odd, which I couldn't solve.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ @julypraiseBy "parity" mean "odd" or "even". We have that $c^2+d^2=0\mod 2$. Since $c^2=c,d^2=d\mod 2$ we have that $c+d=0\mod 2$. This can only happen if $c=d=1\mod 2$ or $c=d=0\mod 2$, that is, either both are odd or both are even. $\endgroup$ – Pedro Tamaroff♦ Oct 17 '13 at 18:13
  • $\begingroup$ Thanks! By the way, you method is also called the proof by infinite descent, right? $\endgroup$ – jachilles Oct 17 '13 at 18:14
  • 1
    $\begingroup$ @julypraise Yes, it is what André mentioned in the comments. I wouldn't know if the above was the proof he suggested, but it should be similar. $\endgroup$ – Pedro Tamaroff♦ Oct 17 '13 at 18:15
  • $\begingroup$ Hey for the odd case, I'm quite stuck. Do I put $ c= 2c'+1$ and likewise for $d$? $\endgroup$ – jachilles Oct 17 '13 at 18:17
  • $\begingroup$ @julypraise Yes. I will try it myself in case it is more complicated. $\endgroup$ – Pedro Tamaroff♦ Oct 17 '13 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.