# How to prove that?

## Oct 24, 2017

See below. (b) is a false statement.

#### Explanation:

(a) $25 = {5}^{2}$. If the number is not divisible by 5, it is not also divisible by 25.
Let's see if ${n}^{2} + n - 9$ is divisible by 5 or not.

For example, if $n$ is divisible by 5, there is an integer $k$ that satisfies $n = 5 k$.
Substitute $n = 5 k$ into ${n}^{2} + n - 9$:
${n}^{2} + n - 9 = {\left(5 k\right)}^{2} + \left(5 k\right) - 9$
$= 25 {k}^{2} + 5 k - 9$
$= 5 \left(5 {k}^{2} + k - 2\right) + 1$.
Thus, the remainder is 1. ${n}^{2} + n - 9$ is not divisible by 5 if $n = 5 k$.

We can express this in congruence.
If n≡0 (mod $5$) then n^2+n-9≡0+0-9≡-9≡1 (mod $5$)

For other cases(n≡1,2,3,4 (mod $5$)), see the table. ${n}^{2} + n - 9$ is not a multiple of $5$ in any cases, neither is it a multiple of $25$.

(b) $49 = {7}^{2}$.
Similarly to (a), find the remainder (mod $7$) first. If n≡0,2,3,4,6 (mod $7$), ${n}^{2} + n - 9$ is a not divisible by $7$ and it is proven that ${n}^{2} + n - 9$ is not a multiple of $49$.

If n≡1 or n≡5 (mod $7$), ${n}^{2} + n - 9$ is a multiple of $7$.
Now let's move to the next stage.

How to show that no natural square $n = {m}^{2}$ satisfies (A),(B)?

(A) n≡1,5 (mod $7$)
(B) n^2+n-9≡0 (mod $49$)

Since m^2≡0,1,2,4 (mod $7$), we can confine to the case n=m^2≡1 (mod $7$). In this case, $m$ must satisfy m≡1,6 (mod $7$) , i.e. m≡1,6,8,13,15,20,22,27,29,34,36,41,43,48 (mod $49$).

Here is a table for $m , n = {m}^{2} , {n}^{2} = {m}^{4} \mathmr{and} {n}^{2} + n - 9 = {m}^{4} + {m}^{2} - 9$. Oops! ${n}^{2} + n - 9 = {m}^{4} + {m}^{2} - 9$ is divisible by $49$ if m≡6,43 (mod $49$) and the statement is false.

For example:
$m = 6 , n = {6}^{2} = 36$
${n}^{2} + n - 9 = {36}^{2} + 36 - 9 = 1323 = 49 \cdot 27$.