Show not possible to find positive whole numbers $m,n$ such that $m^2 − n^2 = 6$.

Question

Show not possible to find positive whole numbers $m,n$ such that $m^2 − n^2 = 6$.

$m^2 − n^2 = 6\implies (m+n)(m-n) = 6; \ m,n$ can be either even or odd.
If $m$ is odd, $m=2k+1, k\ge 1$; else $m=2k$.
Similarly, if $n$ is odd, $n=2l+1, l\ge 1$; else $n=2l$.

There are $4$ possible cases:
(i) $m,n$ both are odd: $(m+n)(m-n)= 4(k+l+1)(k-l)$,
(ii) $m,n$ both are even: $(m+n)(m-n)= 4(k+l)(k-l)$,
(iii) $m,n$ one is odd, other is even: $(m+n)(m-n)= (2(k+l)+1)(2(k-l)\pm 1)$
$= 4(k+l)(k-l)+2(k-l)\pm(2(k+l)+1)$
$= 4(k+l)(k-l) +4k+1, \ \ 4(k+l)(k-l) -4l - 1$

Not possible to pursue further except by proof by contradiction.

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If try by contradiction based proof, need prove that the premise $m^2-n^2=6$ is false:

(i) $(m+n)(m-n)= 4(k+l+1)(k-l)$
$\implies 6 = 4(k^2 + kl + k - kl -l^2 -l)= 4(k^2 -l^2 +k -l) = 4(k-l)(k+l+1)$
$\implies 3 = 2(k+l+1)(k-l)$
The lhs is odd, but the rhs is even.
(ii) $(m+n)(m-n)= 4(k+l)(k-l)$
$\implies 6 = 4(k+l)(k-l)$
$\implies 3 = 2(k+l)(k-l)$
The lhs is odd, but the rhs is even.
(iii) $4(k+l)(k-l) +4k+1$
$\implies 5 = 4(k+l)(k-l)+k)$
The lhs is odd, but the rhs is even.
(iv) $4(k+l)(k-l) -4l - 1$
$\implies 7 = 4(k+l)(k-l)-l)$
The lhs is odd, but the rhs is even.

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However, the same cannot be said for $m^2 - n^2 = 6j, \ m,n,j \ \in \mathbb{N}$.

Answer 1

Note that

$$m^2 =\sum_{k=1}^m(2k-1)\quad \text{and}\quad n^2 =\sum_{k=1}^n(2k-1)$$

Hence,

$$m^2-n^2 = \sum_{k=n+1}^m(2k-1)\stackrel{!}{=}6$$

It follows that $m\leq 3$ and $n+1\geq 2$, but the possible sums are only $3,5,3+5$. So, there is no solution.

Answer 2

There's an easier way to do this. So, you know $(m+n)(m-n) = 6$, and $6 = 2 \cdot 3 = 1 \cdot 6$. If $m,n$ are to be whole numbers, then so must $m+n$ and $m-n$. Thus, we could look at the systems of equations that result by setting factors equal to each other:

$$\left\{\begin{matrix} m+n = 2\\ m-n=3 \end{matrix}\right. \;\;\;\;\; \left\{\begin{matrix} m+n = 3\\ m-n=2 \end{matrix}\right. \;\;\;\;\; \left\{\begin{matrix} m+n = 1\\ m-n=6 \end{matrix}\right. \;\;\;\;\; \left\{\begin{matrix} m+n = 6\\ m-n=1 \end{matrix}\right.$$

Show that, in none of these cases, you do not have both $m,n$ as whole numbers to get a solution.

Answer 3

Another alternative here: $m+n$ and $m-n$ are either both odd or both even i.e. their product is either odd or divisible by $4$ and $6$ is neither.

Answer 4

Seems fine though your proof is long.

As for your question regarding generalization to $6j$.

For the special case where $j$ is odd, we also can't have $m^2-n^2=6j$.

Suppose not, then we have $(m+n)(m-n)=6j$.

If $m+n$ and $m-n$ are both odd, then the LHS is odd but RHS is even.

If $m+n$ and $m-n$ are both even, then LHS is a multiple of $4$, but RHS is not a multiple of $4$.

Answer 5

The answers already added are pretty good in my opinion, especially the one by @trancelocation. However, since nobody has approached this via congruences yet, I wanted to do the same.

We consider the general case, $m^2-n^2 = 6j, m,n,j \ge 1$. It is an elementary fact that any square is congruent to $0,1$ modulo $4$ according to as it is even or odd. Then, rewriting the equation, we have,

$m^2 = n^2 + 6j \equiv n^2 + 2j \ \bmod 4$

If $n$ is even, we must have $2j \equiv 0 \ \bmod 4 \implies j \equiv 0 \ \bmod 2$.

If $n$ is odd, we again have $2j +1\equiv 1\ \bmod 4 \implies j \equiv 0 \ \bmod 2$.

Hence, the general equation only has solutions if $j$ is even.