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

Question

$m^2 − n^2 = 6\implies (m+n)(m-n) = 6; \ m,n$ 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.

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.

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

What are you asking? The question with $6$ or the one with $6j$? — Parcly Taxel, Apr 05 '20 at 05:48
https://math.stackexchange.com/questions/2191177/prove-that-the-equation-x2-y2-2002-has-no-integer-solution/2191179#2191179 — lab bhattacharjee, Apr 05 '20 at 05:49
@ParclyTaxel Just adding a note at end, that it is only possible to prove for $6$; not any multiple thereof. — jiten, Apr 05 '20 at 05:50
@jiten That's because solutions do exist for higher $j$. For example, $12^2-6^2=6\cdot18$. — Parcly Taxel, Apr 05 '20 at 05:52

score 7 · Answer 1 · answered Apr 05 '20 at 05:59

7

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.

answered Apr 05 '20 at 05:59

acat3

12,197

Similar to what @labbhattacharjee suggested. – jiten Apr 05 '20 at 06:01

score 5 · Answer 2 · answered Apr 05 '20 at 05:50

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.

... for example because the first two imply $2m=5$, the last two $2m=7$. — Torsten Schoeneberg, Apr 05 '20 at 07:07

score 3 · Accepted Answer · answered Apr 05 '20 at 06:03

3

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.

answered Apr 05 '20 at 06:03

trancelocation

33,349
1
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Please elaborate your answer more, from the second line. Say, how got that $$m^2-n^2 = \sum_{k=n+1}^m(2k-1)\stackrel{!}{=}6$$ Also, the third line needs more in terms of missing steps. – jiten Apr 05 '20 at 06:09
1

It is a standard result (which can easily be verfied by induction), that $n^2$ can be formed as the sums of the first $n$ odd natural numbers. For example: $1+3 = 2^2$, $1+3+5 = 3^3$ and so on. – trancelocation Apr 05 '20 at 06:13
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In the next step I just form the difference of the sums. For example: $3^2-2^2 = (1+3+5) - (1+3) = 5$. So, you get a sum of positive numbers that is supposed to be equal to $6$. So, the highest summand $2m-1$ is not allowed to exceed $6$. – trancelocation Apr 05 '20 at 06:16
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Since we subtract at least $n=1=2\cdot 1-1$, the index $k$ of the final sum has to start at least at $k=n+1=2$. – trancelocation Apr 05 '20 at 06:18
Thanks a lot for this approach. – jiten Apr 05 '20 at 06:20
1

You are welcome :-). – trancelocation Apr 05 '20 at 06:22

Siong Thye Goh · Answer 4 · 2020-04-05T06:55:33.127

2

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 $4$.

edited Apr 05 '20 at 06:55

answered Apr 05 '20 at 06:47

Siong Thye Goh

153,832

Please see my post at : https://math.stackexchange.com/q/3613426/424260. – jiten Apr 07 '20 at 03:40
Please help with my post stated in the last comment. You can help in chatroom alternatively. – jiten Apr 07 '20 at 04:23
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I am working on some other stuff, I might post an answer if I have the time and if it is still not resolved later – Siong Thye Goh Apr 07 '20 at 04:24
Any sort of help is not coming. – jiten Apr 07 '20 at 08:03
Sorry late, but is based on implicit proof of $4 \nmid 6j, j \in { 2m+1}, m\in \mathbb{N}$$ \implies 2 \nmid 3j, j \in { 2m+1}, m\in \mathbb{N} $. – jiten Mar 03 '21 at 01:05

score 1 · Answer 5 · answered Apr 05 '20 at 07:45

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.

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

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