Integral Solutions

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Brut3Forc3

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Brut3Forc3

#1 h Apr 4, 2010, 2:41 AM • 4 Y

Y by ahmedosama, Adventure10, Mango247, and 1 other user

Determine all integral solutions of $a^2+b^2+c^2=a^2b^2.$

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basketball9

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basketball9

#2 h Apr 4, 2010, 3:40 AM • 2 Y

Y by Adventure10, Mango247

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xpmath

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xpmath

#3 h Apr 4, 2010, 3:48 AM • 2 Y

Y by Adventure10, Mango247

solution

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harekrishna

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harekrishna

#4 h May 17, 2010, 11:40 AM • 2 Y

Y by Adventure10, Mango247

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

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Arrange your tan

970 posts

Arrange your tan

#5 h May 17, 2010, 8:25 PM • 4 Y

Y by CircleGeometryGang, sabkx, Adventure10, Mango247

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.)

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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.

This post has been edited 1 time. Last edited by Arrange your tan, May 18, 2010, 3:28 AM

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andersonw

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andersonw

#6 h May 18, 2010, 12:58 AM • 2 Y

Y by Adventure10, Mango247

@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.

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varunrocks

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varunrocks

#7 h May 18, 2010, 1:09 AM • 1 Y

Y by Adventure10

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.

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AIME15

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AIME15

#8 h May 18, 2010, 2:00 AM • 2 Y

Y by Adventure10, Mango247

Actually, the equation is

$\begin{align*} a^2+b^2+c^2 & = a^2b^2 \\ 4a_1^2+4b_1^2+4c_1^2 &= 16a_1^2b_1^2 \\ a_1^2+b_1^2+c_1^2&=4a_1^2b_1^2. \end{align*}$

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xpmath

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xpmath

#9 h May 18, 2010, 2:07 AM • 2 Y

Y by Adventure10, Mango247

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

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varunrocks

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varunrocks

#10 h May 18, 2010, 3:29 AM • 2 Y

Y by Adventure10, Mango247

Sorry, I also have typoed!

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Zhero

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Zhero

#11 h May 18, 2010, 3:29 AM • 7 Y

Y by maXplanK, CircleGeometryGang, Adventure10, Mango247, and 3 other users

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.

A rigorized version of xpmath's solution

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SodaKing1

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SodaKing1

#12 h Aug 13, 2013, 7:32 PM • 2 Y

Y by Adventure10, Mango247

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.

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NT2048

374 posts

NT2048

#13 h Aug 13, 2013, 9:18 PM • 2 Y

Y by Adventure10, Mango247

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.

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fighter

507 posts

fighter

#14 h Sep 9, 2016, 11:52 AM • 2 Y

Y by Adventure10, Mango247

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

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AllenWang314

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AllenWang314

#15 h Mar 13, 2018, 8:42 PM • 1 Y

Y by Adventure10

Where does this go wrong?

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djmathman

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djmathman

#17 h Mar 13, 2018, 9:05 PM • 1 Y

Y by Adventure10

AllenWang314 wrote:

snippit

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.)

This post has been edited 1 time. Last edited by djmathman, Mar 13, 2018, 9:05 PM

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AllenWang314

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AllenWang314

#18 h Mar 14, 2018, 3:21 AM • 2 Y

Y by Adventure10, Mango247

explanation

Is this correct?

This post has been edited 1 time. Last edited by AllenWang314, Mar 14, 2018, 3:22 AM

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djmathman

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djmathman

#19 h Mar 14, 2018, 5:08 PM • 1 Y

Y by Adventure10

Quote:

In particular, one of these is $3$ mod $4$ and must be divisible by a prime $3$ mod $4$ .

Not if $a=0$ !

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Deligne

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Deligne

#20 h Feb 29, 2020, 4:18 PM

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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 $a=b=c=0$ .

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huashiliao2020

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huashiliao2020

#21 h Apr 14, 2023, 4:15 AM

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xpmath wrote:

solution

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.

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HamstPan38825

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HamstPan38825

#22 h Apr 21, 2023, 11:06 PM

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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.

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Seungjun_Lee

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Seungjun_Lee

#23 h Apr 22, 2023, 10:25 AM

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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 $a=c=0$

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cj13609517288

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cj13609517288

#24 h Apr 25, 2023, 3:57 PM

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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 $(0,0,0)$ .

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p.lazarov06

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p.lazarov06

#25 h May 2, 2023, 11:23 AM

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$\begin{align*} a^2+b^2+c^2&=a^2b^2\\ a^2b^2-a^2-b^2+1&=c^2+1\\ (a^2-1)(b^2-1)&=c^2+1\\ \end{align*}$
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:

$(a;b;c)=(0;0;0)$

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surpidism.

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surpidism.

#26 h May 20, 2023, 5:52 AM

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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 $a = b = c = 0$ .

This post has been edited 1 time. Last edited by surpidism., May 20, 2023, 5:56 AM

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Pyramix

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Pyramix

#27 h May 20, 2023, 1:34 PM

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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 $\boxed{(a,b,c)=(0,0,0)}$ .

This post has been edited 1 time. Last edited by Pyramix, May 20, 2023, 1:35 PM
Reason: Unnecessary bounds removed.

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shendrew7

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shendrew7

#28 h May 2, 2024, 4:33 AM

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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 $\boxed{(0,0,0)}$ . $\blacksquare$

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Mr.Sharkman

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Mr.Sharkman

#29 h May 14, 2024, 11:39 PM

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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 $a=b=0.$

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MuradSafarli

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MuradSafarli

#30 h Mar 31, 2025, 1:36 PM

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(a,b,c)=(0,0,0) clearly works,otherwise
"Let O be odd and E be even. The following combinations (a,b,c) – (O,E,E), (E,O,E), (E,E,O), (O,O,E), (O,E,O), (E,O,O), (O,O,O), (E,E,E) easily yield contradictions modulo 4."

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cursed_tangent1434

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cursed_tangent1434

#31 h Apr 28, 2025, 2:55 PM

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We claim that the only triple of solutions is $(a,b,c)=(0,0,0)$ which clearly works. We now show that this is the only solution. First note that if $(a,b,c)$ is a triple of solutions $|a|,|b| \ne 1$ since this reduces the equation to $c^2+1=0$ which has no real solutions. We also observe that if $(a,b,c)$ is a triple of solutions so are all triples of the form $(\pm a,\pm b,\pm c)$ . Hence it suffices to solve the equation over positive integers. We rearrange the equation to,
$c^2+1=(ab)^2-a^2-b^2+1= (a^2-1)(b^2-1)=(a-1)(a+1)(b-1)(b+1)$ If $a$ is odd, one of $a-1$ and $a+1$ is divisible by $4$ and thus, $4\mid c^2+1$ which is impossible for all positive integers $c$ . If $a$ is even, one of $a-1$ and $a+1$ is $3 \pmod{4}$ . Thus, there exists a prime $p \equiv 3 \pmod{4}$ which divides the right-hand side. But then, $p \mid c^2+1$ which is impossible by Fermat's Christmas theorem. Hence, $a$ cannot take any positive integral value, implying that all solutions are of the claimed forms.

This post has been edited 1 time. Last edited by cursed_tangent1434, Apr 29, 2025, 9:18 AM
Reason: fakesolve

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