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.
Is $7m^2-3n^2$ a perfect square?
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$\begingroup$
algebra-precalculus
elementary-number-theory
square-numbers
sums-of-squares
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2What made you think this is always a square? Trying a few examples at random...it's hardly ever a square. – 2017-01-29
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0@lulu what if $m,n\gt1$? – 2017-01-29
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0Try 3,2 for example. Or 5,2 or 6,2 or 7,2. Really, the thing is hardly ever a square. – 2017-01-29
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0@lulu thanks, by the way – 2017-01-29
3 Answers
6
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.
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0ok, what if $m,n\gt1$? – 2017-01-29
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0@vidyarthi Does this answer your question? – 2017-01-29
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0thanks, that answers the question succintly! – 2017-01-29
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1@vidyarthi Be sure to check all cases the next time you make a claim. Good luck. :) Could you accept my answer. – 2017-01-29
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1S.C.B, in case of interest, the times when a square does occur can be parametrized, similar to the way we get Pythagorean triples, but requiring more than one bunch of formulas. In order to get every possible combination of $\pm$ signs, a few more very similar formulas would be required. – 2017-01-29
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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:
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$$ x = 3 u^2 + v^2 , $$ $$ y = |3 u^2 + 4 u v - v^2|, $$ $$ z = |6 u^2 - 6 u v - 2 v^2|. $$
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$$ x = 2 u^2 + 2uv + 2 v^2 , $$ $$ y = |3 u^2 + 4 u v - v^2|, $$ $$ z = | -u^2 +8 u v +5 v^2|. $$
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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
0
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.