An interesting LL problem of IMO 1971

[nafistiham]

Notice: $8^3-7^3=169=13^2$ and $13=2^2+3^2$.
Prove that if the difference between two consecutive cubes is a square, then it is the square of the sum of two consecutive squares.

[zadid xcalibured]

not square of the sum.sum of the square

[nafistiham]

not square of the sum.sum of the square

I also though that.but, look carefully.It is OK

[Fahim Shahriar]

For some integers m and n, we have \[(m + 1)^{3} - m^{3}\ = 3m^{2}+3m+1=n^{^{2}}\] .
So, \[12m^{2}+12m+4 = 4n^{_{2}}\], from which we can get \[3(2m+1)^{2}= (2n+1)(2n-1)\].

Now (2n+1) & (2n-1) are RELATIVELY PRIME. Since both are odd, their greatest common divisor must be 1.

Then, we have two possible cases.
1. \[2n - 1 = 3a^{2}\], \[2n + 1 = b^{2}\]

2. \[2n - 1 = a^{2}\], \[2n + 1 = 3b^{2}\]

For the FIRST case, \[b^{2}=3a^{2}+2\]
I know this equation is impossible, but don't know the reason properly.


For the second case, as 2n-1 is an odd number, setting a = 2k + 1 .... \[ 2n-1=(2k+1)^{2}=4k^{2}+4k+1\]
Hence, \[ 2n=4k^{2}+4k+2 = 2[k^{2}+(k+1)^{2}]\]
So, n= \[k^{2}+(k+1)^{2}\]

{Proved}

[Masum]

For the FIRST case, \[b^{2}=3a^{2}+2\]
I know this equation is impossible, but don't know the reason properly.
{Proved}

Recall that \[x^2\equiv0,1\pmod3\]