Is $7m^2-3n^2$ a perfect square?

Question

Is $7m^2-3n^2$ a perfect square for all positive integers $m,n$?

I tried using double induction, but failed. Any other approach? By the way, is this related to fermat's theorem on representation of an integer by sums of squares? Thanks beforehand.

Answer 1

Counterexample: $m=3, n=1$. Then $$7m^2-3n^2=63-3=60$$

$60$ is not a perfect square. Thus, your claim is false. It is not a perfect square for all $m,n$.

Counterexample 2 (As a reply to the comment)

Take $m=3k, n=k$ where $k$ is any integer. Then $$7m^2-3n^2=60k^2$$ $60$ is not a square. So $60k^2$ is not a square.

Answer 2

The occurrences of $$ 7 x^2 - 3 y^2 = z^2 $$ with $\gcd(x,y,z) = 1$ come in two parametrized families, depending on whether $z$ or $x$ is even:

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

$$ x = 3 u^2 + v^2 , $$ $$ y = |3 u^2 + 4 u v - v^2|, $$ $$ z = |6 u^2 - 6 u v - 2 v^2|. $$

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

$$ x = 2 u^2 + 2uv + 2 v^2 , $$ $$ y = |3 u^2 + 4 u v - v^2|, $$ $$ z = | -u^2 +8 u v +5 v^2|. $$

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

In this printout I did not use any absolute value signs. Doesn't matter.

    x    y    z             u    v
    1   -1   -2             0    1
    2   -1    5             0    1
    2    3   -1             1    0
    7   -9   10             1   -2
   13   19   10             2    1
   13    3   34             2   -1
   14   -1  -37             3   -2
   14   19   17             2    1
   19  -29   -2             1   -4
   19    3  -50             1    4
   26  -29   47             1   -4
   26   31  -43             4   -1
   26  -37  -25             3   -4
   26   -9  -67             4   -3
   31   -1   82             3   -2
   31   47   10             3    2
   37   27  -86             2    5
   37  -53   34             2   -5
   38   27   89             2    3
   38   31  -85             5   -2
   38   47   59             3    2
   38  -53   41             2   -5
   43  -37   94             3   -4
   43   59  -50             3    4
   49   31  118             4   -1
   61   19 -158             2    7
   61  -93   10             2   -7
   62  -37 -151             6   -5
   62  -57  131             1   -6
   62  -81  -85             5   -6
   62   83  -79             6   -1
   67  -29 -170             1    8
   67  -93  -74             1   -8
   73  103  -74             4    5
   73  -57  166             4   -5
   74 -113    5             4   -7
   74   19 -193             7   -4
   74   59  167             3    4
   74   87  125             4    3
   79  111   82             5    2
   79   31  202             5   -2
   86 -109 -127             6   -7
   86  131   17             6    1
   86  -57 -205             7   -6
   86   -9  227             1    6
   91  139   -2             5    4
   91   59 -218             3    8
   97  111 -170             4    7
   97 -113  166             4   -7
   98 -149  -25             5   -8
   98   87 -211             8   -3

Answer 3

If $m=1$ and $n>1$ then $7m^{2}-3n^{2}<0,$ so it can't be a perfect square as perfect squares are non-negative.

More generally, since $n\mapsto n^{2}$ is not bounded above, for each value of $m$ there exists a positive integer $N$ such that $7m^{2}-3n^{2}<0$ for all $n>N,$ so that for each value of $m$ there are infinitely many choices of $n$ such that the expression is not a perfect square.