# Thread: Stuck on a proof

1. ## Stuck on a proof

I'm stuck on the final step in these two proofs:
1) For every natural number N that leaves remainder of 1 when divided by 5, there's a n^2 that also leaves remainder of 1 when divided by 5

n = 5k + 1 where k, an integer, is the quotient:
Then n^2 = (5k + 1)^2
n^2 = 25k^2 + 10k + 1
= 5(5k^2 + 2k) + 1

Stuck here...

let j be 5k^2 + 2k
if k is an integer, then 5k^2 + 2k is also an integer
n^2 = 5j + 1, where j is an integer quotient

And the very similar problem...

2) For every natural number N that leaves remainder of 4 when divided by 5, there's a n^2 that leaves remainder of 1 when divided by 5

n = 5j + 4 where j, an integer, is the quotiet:
Then n^2 = (5j + 4)^2
n^2 = 25j^2 + 40j + 16
= 25j^2 + 40j + 15 + 1
= 5(5j^2 + 8j + 3) + 1
= 5(j + 1)(5j + 3) + 1

Stuck here...

Thanks for any help.

2. hi selena,

you've proved it! There's nothing more to do... you might want to think about how you present the proof. The implication is:
$\displaystyle (\forall n)(\exists j, k)(n = 5j + 1) \Rightarrow (n^2 = 5k + 1); n, j, k \in \mathbb{N}$

and then you use your proof to show:
$\displaystyle n^2 = 5(5j^2 + 2k) + 1$

and notice that $\displaystyle k = 5j^2 + 2j \Rightarrow k \in \mathbb{N}$. You normally don't have to show that any two integers add or multiply to give another integer.