I am not sure about xpmath's solution. Shouldn't $ {a_1} ^2 + {b_1}^2 + { c_1} ^ 2 = 4 (a_1 b_1)^2 $

basketball9 wrote: We divide both sides by $ a^2b^2$.... [1] So we get $ 1/a^2 + 1/b^2 + c^2 = 1$ [2] so $ 1/a^2 + 1/b^2 = 1 - c^2/a^2b^2$ [1] a) The third term should be $\frac{c^2}{a^2b^2}$. b) While this equation would have been mathematically correct (see [1] a above), it is misleading, because the direct next step after dividing by $a^2b^2$, but befoe reordering terms, would be $\frac {1}{b^2} + \frac {1}{a^2} + \frac {c^2}{a^2b^2} = 1$. [2] If you type it out horizontally, it has to have grouping symbols around the denominator such as $\frac{1}{a^2} + \frac{1}{b^2} = 1 - c^2/(a^2b^2)$ Otherwise, type out the fractions in a vertical style and avoid the needed use of grouping symbols: $\frac{1}{a^2} + \frac {1}{b^2} = 1 - \frac{c^2}{a^2b^2}$ (This post is being neutral towards the correctness of your overall method.) - - -- - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - harekrishna wrote: I don't know about xpmath's solution. Shouldn't ${a_1}^2 + {b_1}^2 + {c_1}^2 = 4(a_1b_1)^2?$ . . . EDITED Yes, it should be the equivalent to that.

@basketball: I don't see how that shows that c needs to be 0. As long as c<ab, the right hand side is still positive. xpmath's solution still works, because the right side will always be 0 mod 4, so all of the terms on the left side need to be 0 mod 2 as well (because a perfect square is only 0 or 1 mod 4, of course), and you can still perform the infinite descent. The coefficient of the right hand side does grow larger, but it is still always 0 mod 4.

I also do not understand Basketball9's logic. @harekrishna: xpmath got that 2a_1=a, 2b_1=b, 2c_1=c so we get: 4(a_1)^2+4(b_1)^2+4(c_1)^2=4(a_1)(b_1) Dividing through by 4, we get: (a_1)^2+(b_1)^2+(c_1)^2=(a_1)(b_1) Which is correcy.

Actually, the equation is \begin{align*} a^2+b^2+c^2 & = a^2b^2 \\ 4a_1^2+4b_1^2+4c_1^2 &= 16a_1^2b_1^2 \\ a_1^2+b_1^2+c_1^2&=4a_1^2b_1^2. \end{align*}

Hmm must've typoed, sorry. The idea is still the same though, like Anderson said.

Sorry, I also have typoed!

Rearrange this as $c^2 + 1^2 = (a^2 - 1)(b^2 - 1)$. By looking at this mod 4, it can easily be seen that $c$ must be even. It is well-known that if $p | c^2 + 1^2$, then $p = 2$, $p \equiv 1 \pmod{4}$, or $p | c, 1$. Since $c^2 + 1^2$ is odd and no prime divides 1, we must have that all prime factors of $c^2 + 1^2$ are congruent to 1 modulo 4. Hence, the product of the prime factors of $a^2 - 1$ must be conrguent to 1 modulo 4. But $a^2 - 1 \equiv 0, 3 \pmod{4}$, so we have a contradiction. Hence, there are no nontrivial solutions.

Sorry for bringing this back up but could someone explain what infinite descent means? I understand how the first part of the solution but not the descent part. Also if someone could pm me maybe a link to somewhere that explains the concept well.

It's not really a fancy concept or anything. We start with this equation: \[ a^2+b^2+c^2=a^2b^2.\] We find that $a, b, c$ are even and we let $ a=2a_1, b=2b_1, c=2c_1$. We find $ {a_1}^2+{b_1}^2+{c_1}^2=4{a_1}^2{b_1}^2$. Using another argument, we can show that $a_1, b_1, c_1$ are even. So let $ a_1=2a_2, b_1=2b_2, c_1=2c_2$. Then we sub this into the equation and find that $a_2, b_2, c_2$ are even. And we can repeat this argument an infinite number of times to get that $a_3, a_4 ... a_i ...$ are all even and it would imply that $a, b, c$ have an infinite number of factors of 2. But this is impossible, so solutions can't exist.

lemma-1: a ,b ,c are even proof: using mod 4; now, a = 2*a', b = 2*b', c = 2*c' then, a'^2 + b'^2 + c'^2 = (a'^2)*(b'^2) which is a recurrence; so, the solutions are a = b = c = 0

AllenWang314 wrote:

This statement is the first place where your solution breaks, and it feels really fishy to me. In particular, I have no idea how non-QR-ness of two integers imply that they both must be even. (Also keep in mind that any two non-QRs mod a prime must multiply to a QR.)

Is this correct?

Quote: In particular, one of these is $3$ mod $4$ and must be divisible by a prime $3$ mod $4$. Not if $a=0$!

Assume without loss of generality that $a\geqslant b$. Then $a^2 b^2 =a^2 +b^2 +c^2 \leqslant 2a^2 +c^2$. Therefore $c^2 \geqslant a^2(b^2- 2)$. Since $b^2 -2$ is not a perfect square, we have that $c^2 \neq a^2 (b^2-2)$. Similarly $c^2 \neq a^2 (b^2-1)$. Hence $c^2 \geqslant a^2 b^2$. It follows that $a=b=c=0$.

xpmath wrote:

Yup, that's what I did, except that a_1^2+b_1^2+c_1^2=4a_1^2b_1^2, which again is 0 mod 4 on RHS, so LHS must be 0 mod 2 for each of a_1, b_1, and c_1 to have their sum of their squares by 0 mod 4. Also, a bit of an explanation. If both a and b are odd, then the RHS is 1 mod 4, and LHS is 1 (a^2) + 1 (b^2) + c^2 mod 4, must be 1 mod 4, absurd, because this implies c^2 must be 3 mod 4. Now if at least of one of a and b is even, then RHS is 0 mod 4, LHS is 0 mod 4 + b^2 + c^2, or b^2+c^2 must be 0 mod 4, absurd (2+2, 1+3 all impossible) unless both b and c are 0 mod 2. Now a and b must both be 0 mod 2, meaning RHS is even and c must also be even for even + even + even = even.

Write the given equation as $$(a^2-1)(b^2-1) = c^2+1.$$By Fermat Christmas, divisors of the RHS must be $2$ or 1 mod $4$. On the other hand, the factors on the LHS are either multiples of $4$ or $3$ mod $4$, unless $(a, b, c) = (0, 0, 0)$. Indeed this is the sole solution.

Since $c \le ab$ Let $c = ab - k$ $a^2 + b^2 + k^2 - 2abk = 0$ Let $(k,a,b)$ is the solution with the minimum $k +a+b$ Let $k \ge a \ge b>0$ Let $f(x) = x^2 - 2abx + a^2 + b^2$ Let two solutions of $f(x) =0$ as $x_1=k$ and $x_2$ $x_2 \in \mathbb{Z}$ $f(a) \ge 0$ Then $3a^2 \ge 2a^2 + b^2 \ge 2a^2b$ Then $b= 1$ $(a-k)^2 + 1 = 0$ Contradiction Then $b=0$ and $a=c=0$

Take mod $4$. This is either $0+0+0=0$ or $0+0+1=1$. But the second one requires $a$ and $b$ to be both odd, contradiction. The first one falls to infinite descent and has solution $(0,0,0)$.

\begin{align*} a^2+b^2+c^2&=a^2b^2\\ a^2b^2-a^2-b^2+1&=c^2+1\\ (a^2-1)(b^2-1)&=c^2+1\\ \end{align*} It's clear that $a$ and $b$ can't be both odd at the same time, because $c^2+1\not\equiv\pmod{4}$. Now if both $a$ and $b$ are at least $2$, so $(a^2-1)(b^2-1)$ will have a prime divisor $p=4k+3$, and by Fermat's Christmas Theorem $p$ will divide both $c$ and $1$ which is impossible. And now by bashing the small values of $a$ and $b$ we get the only solution to be: \[(a;b;c)=(0;0;0)\]

Note that perfect squares are $0$ or $1$ $(mod\ 4)$. $a^2b^2 \equiv 1(mod\ 4)$ is not possible as this lead both $a^2$ and $b^2$ are $1(mod\ 4)$. So, $LHS \not\equiv RHS$ $(mod\ 4)$ Thus, $a^2b^2 \equiv 0 (mod\ 4)$ and $a$, $b$, $c$ are even. Let $a = 2a_1$, $b = 2b_1$, $c = 2c_1$ and substitute in the original equation. \begin{align*} 4a_1 + 4b_1 + 4c_1 = 16a_1^2b_1^2\\ a_1 + b_1 + c_1 = 4a_1^2b_1^2\ \end{align*}Again $a_1$, $b_1$, $c_1$ are even. Let $a_1 = 2a_2$, $b_1 = 2b_2$, $b_1 = 2c_2$ and we find that $a_2$, $b_2$, $c_2$ are also even. If we repeat this argument infinitely many times, we see that $a$, $b$, $c$ can be divided by $2$ infinitely many times, which is only possible when $a = b = c = 0$.

Brut3Forc3 wrote: Determine all integral solutions of \[ a^2+b^2+c^2=a^2b^2.\] If $a=0$ or $b=0$ then all three must be 0. If $c=0$ then $(a^2-1)(b^2-1)=1$, which has only solution $a=b=c=0$. Now, take $a,b,c>0$. Clearly, $a=b=c$ is not possible unless $a=b=c=0$. Also, $a,b>1$. (i.e., $a^2,b^2\geq4$). Note that $c^2+1=(a^2-1)(b^2-1)$. So, if an odd prime $p\mid(a^2-1)$, then $p\mid(c^2+1)$. This means $p\equiv1\pmod{4}$. It follows that if $a$ is even, then all divisors of $a^2-1$ are $\equiv1\pmod{4}$. So, $a^2-1\equiv1\pmod{4}$, a contradiction. It follows that $a$ must be odd. Similarly, $b$ must also be odd. But then $c^2\equiv a^2b^2-a^2-b^2\equiv1-1-1\equiv3\pmod{4}$, a contradiction. Therefore, the only integer solution to the given equation is $\boxed{(a,b,c)=(0,0,0)}$.

Rearranging, we find \[c^2+1 = (a^2-1)(b^2-1).\] Fermat's Christmas Theorem tells us all factors of the LHS with magnitude greater than 1 must be 1 or 2 modulo 4, contradiction. Thus $c^2+1 \leq 1 \implies c=0$, giving the only solution $\boxed{(0,0,0)}$. $\blacksquare$

Simplifying our expression, we get $$(a^{2}-1)(b^{2}-1) = c^{2}+1. $$If $c$ is odd, then $$c^{2}+1 \equiv 2 \pmod 4. $$But, if $2|(a^{2}-1)(b^{2}-1), $ then $8|(a^{2}-1)(b^{2}-1),$ so $c$ is even. Now, if $p|c^{2}+1,$ we have that $c^{2} \equiv -1 \pmod p, $ so $$\left(\frac{-1}{p} \right) = 1.$$Hence, $p \equiv 1 \pmod 4.$ Now, $a^{2}-1$ is odd, so $a$ is even. Thus, one of $a-1$ or $a+1$ is $1 \pmod 4,$ and the other is $3 \pmod 4.$ But, these are divisors of $c^{2}+1,$ and all divisors of $c^{2}+1$ are $1 \pmod 4,$ so this is impossible. Thus, there are no solutions, except for when $c=0,$ since then there are no prime factors $p,$ and when $a=b=0.$