Let a = 3 ^ m * a1, b = 3 ^ l * b1, c = 3 ^ k * c1, n = 3 ^ p * n1, where a1, b1, c1, n1 are not divisible by 3. Then LS simplifies and factors to: 3 * 2 * ( 2a1 * 3 ^ (2m + 1) + b1 * 3 ^ (2l + 1) + c1 * 3 ^2k). Because 3 doesn't divide a1, b1, c1, the left side can be expressed as d * 3 ^ (2i + 1), i >= 0, d isn't divisible by 3. RS is 5n1 * 3 ^ 2p and because 3 doesn't divide n1, we get RS = e * 3 ^ 2j, j >= 0, e isn't divisible by 3. But LS = RS , so because 3 doesn't divide e or d, we must have 3 ^ 2p = 3 ^ (2i + 1), which is impossible, or d = e = 0. So, d = e = 0, so a = b = c = n = 0 as required.

SORI!!! i got lost in your solution how did you get the simplification of the left side???

Micchi wrote: SORI!!! i got lost in your solution how did you get the simplification of the left side??? Let a = 3 ^ m * a1, b = 3 ^ l * b1, c = 3 ^ k * c1, n = 3 ^ p * n1, where a1, b1, c1, n1 are not divisible by 3. Then: 6(6*a^2+3*b^2+c^2) =3 * 2 * (2 * 3 * a1 ^ 2 * 3 ^ 2m + 3 * b1 ^ 2 * 3 ^ 2l + c1 ^ 2 * 3 ^ 2k) =3 * 2 * ( 2 * a1 ^ 2 * 3 ^ (2m + 1) + b1 ^ 2 * 3 ^ (2l + 1) + c1 ^ 2 * 3 ^2k) Is that good enough?

Hey could you please use $\LaTeX$? It's not very difficult to learn and then maybe I'll at least be able to read your solution without getting a head ache

rem wrote: Because 3 doesn't divide a1, b1, c1, the left side can be expressed as d * 3 ^ (2i + 1), i >= 0, d isn't divisible by 3. I don't understand how this come from?? Can somebody explain this more clearly? Thanks!

Hmmm, I'm not sure about your solution REM. You don't necessarily have an odd number of powers of 3 on the LHS. My first inclination was to do an infinite descent on factors of 3 too, but it doesn't work immediately like that. What if for example that a and b have no factors of 3 and c has one factor of three, then there are two on the left which is fine.

Here I got some solution from the APMO website: The left side is divisible by 3, so 3 must divide n. So $5n^{2}-36a^{2}-18b^{2}$ is divisible by 9, so 3 must divide c. We can now divide out the factor 9 to get: $4a^{2}+2b^{2}+6d^{2}=5m^{2}$. Now take a, b, d, m to be the solution with the smallest m, and consider residues mod 16. Squares = 0, 1, 4, or 9 mod 16. Clearly m is even so $5m^{2}$ = 0 or 4 mod 16. Similarly, $4a^{2}$ = 0 or 4 mod 16. Hence $2b^{2}+6d^{2}$ = 0, 4 or 12 mod 16. But $2b^{2}$ = 0, 2 or 8 mod 16 and $6d^{2}$ = 0, 6 or 8 mod 16. Hence $2b^{2}+6d^{2}$ = 0, 2, 6, 8, 10 or 14 mod 16. So it must be 0. So b and d are both even. So a cannot be even, otherwise m/2, a/2, b/2, d/2 would be a solution with smaller m/2 < m. So we can divide out the factor 4 and get: $a^{2}+2e^{2}+6f^{2}=5k^{2}$ with $a$ odd. Hence k is also odd. So $5k^{2}-a^{2}$ = 4 or 12 mod 16. But we have just seen that $2e^{2}+6 f^{2}$ cannot be 4 or 12 mod 16. So there are no solutions.

Doh, shouldn't have given up at mod 8. I really dislike the diophantine equations where all you do is take a bunch of clever mods and it falls apart.

Well what can I say, I don't know latex ... But analize the power parity

First note that $6|n$, so for some $n_1$ we have $6a^2+3b^2+c^2=30n_1^2$. From here we have $3|c$, so \[2a^2+b^2+3c_1^2=10n_1^2\]for some $c_1$, whence $b\equiv c_1\pmod2$. If both are odd, then taking modulo 8 we get \[2a^2-2n_1^2\equiv4\pmod8\implies a^2-n_1^2\equiv2\pmod4,\]a contradiction. So both are even, and integers $b_1,c_2$ exist such that $a^2+2b_1^2+6c_2^2=5n_1^2$. Now $a,n_1$ have the same parity, and like before both cannot be odd or else \[b_1^2-c_2^2\equiv b_1^2+3c_2^2\equiv2\pmod4.\]So both are even, and $a_1,n_2$ exist with \[2a_1^2+b_1^2+3c_2^2=10n_2^2.\]By the method of infinite descent we must have $(a,b,c,n)=(0,0,0,0)$.

This is an APMO-1989 problem.

math154 wrote: If both are odd, then taking modulo 8 we get \[2a^2-2n_1^2\equiv4\pmod8\implies a^2-n_1^2\equiv2\pmod4,\]a contradiction. Could someone explain this step more? I don't understand how the initial mod was obtained.

NuncChaos wrote: math154 wrote: If both are odd, then taking modulo 8 we get \[2a^2-2n_1^2\equiv4\pmod8\implies a^2-n_1^2\equiv2\pmod4,\]a contradiction. Could someone explain this step more? I don't understand how the initial mod was obtained. If $ac\equiv bxc\mod n,a\equiv b\mod \frac n {gcd (n,c)}$

The L.H.S is $36a^2+18b^2+6c^2$ The right hand side is divisible by 5. Now , $36a^2+18b^2+6c^2\equiv {a^2+3b^2+c^2} (mod 5) $ and $a^2+3b^2+c^2\equiv 0 (mod5)$.....................(1) We know $x^2\equiv 0,1,-1 (mod 5)$ It is not possible to show that sum of them are divisible by 5 according to the equation (1), a contradiction So the only answer is $a = b = c = n = 0$

@above its not that simple if $a^2 \equiv 1$ and $c^2 \equiv -1 $and $b \equiv 0 $ then there is no problem mod $5$

Infinite descent???

Note that $6|n$ and $3|c$, so subbing $n=6n'$ and $c=3c'$ yields $b^2+2a^2+3c'^2 = 10n'^2$. Now we have $b^2+3c'^2\equiv 0$ or $\pm 2\pmod{8}$ (the latter is impossible), so $2|b$ and $2|c'$, and subbing $b=b'$ and $c=2c''$ yields $a^2+2b'^2+6c''^2=5n'^2$. This time we get $5n'^2-a^2\equiv 0$ or $\pm 2\pmod{8}$ (the latter is impossible) so $2|a$ and $2|n'$ and we win by infinite descent.

Note that $6 \mid n$, so substitute $n = 6n_1$. We get \[6a^2+3b^2+c^2 = 30n_1^2,\] so $3 \mid c$. Substutituing $c = 3c_1$ gives \[b^2+3c_1^2=10n_1^2-2a^2.\] Analyze this equation modulo $8$. The terms on the LHS are either $0$ or $2$, so the RHS can only be $0$ or $\pm 2$. Testing out all cases on the LHS show the latter to be impossible, and we must have $2 \mid b,c_1$. Plug in $b = 2b_1$ and $c_1 = 2c_2$ to get \[2b_1^2+6c_2^2 = 5n_1^2-a^2.\] A similar argument on $n_1$ and $a$ yield that they must be even. Hence, we plug in $n_1=2n_2$ and $a=2a_1$: \[b_1^2+3c_2^2 = 10n_2^2-2a_1^2,\] upon which we use infinite descent to finish. $\square$