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?

  • $\begingroup$ 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

$${ \left( \frac { a }{ b } \right) }^{ 2 }+1={ \left( \frac { c }{ b } \right) }^{ 2 }\\ \frac { c }{ b } +\frac { a }{ b } =k\\ \frac { c }{ b } -\frac { a }{ b } =\frac { 1 }{ k } \\ \frac { c }{ b } =\frac { { k }^{ 2 }+1 }{ 2k } \\ \frac { a }{ b } =\frac { { k }^{ 2 }-1 }{ 2k } \\ $$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?

  • $\begingroup$ 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

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.


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:

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


  • $\begingroup$ 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

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.

  • 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
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    $\begingroup$ @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

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