Is there any way to prove that if $4y=4x+13k$ then $y=x+13l$?

Question

I am doing a true/false problem, and it says if $4y=4x+13k$ then $y=x+13l$. I couldn’t find any counter example so I suppose the statement is true but could not prove the above equation equivalent to $y=x+13l$.

So is the problem true or false?

Thank you

Answer 1

Alt. hint:   using that $\,3 \cdot 4 = 13 - 1\,$, multiplying by $3$ gives:

$$12y=12x+ 3 \cdot 13k \iff 13y - y = 13 x - x + 3 \cdot 13 k \iff y = x - 13 \cdot (x -y + 3k)$$

Answer 2

Given that $$4y\equiv 4x\pmod {13}$$ Since $\gcd(4,13)=1$ we have that $$y\equiv x\pmod {13}\implies y=x+13k\qquad \forall k\in\mathbb{N}$$

Answer 3

We have $$4x\equiv4y\pmod{13}\implies4x-4y=13k$$ for some $k\in\mathbb{Z}$. LHS is divisible by $4$ so $4\mid k$ for the equality to hold. So let $k=4l$, $l\in\mathbb{Z}$. Then $$4x-4y=13(4l)\implies x-y=13l\implies x\equiv y\pmod{13}$$ as required.

Answer 4

Here are a couple of observations.

(a) multiply by $10$ to obtain $40y=40x+130k$ or $y=x+13(3x-3y+10k)$. Why choose $10$? - Well because $10\times 4 \equiv 1 \bmod 13$, and the multiplicative inverse of $3$ is guaranteed modulo $13$ because $4$ and $13$ are coprime.

(@dxiv's answer uses $3\times 4 \equiv -1 \bmod 13$ in a similar way - note that $10+3=13$ so the multipliers are related)

(b) consider the prime factorisation of $k$ - it must be even, otherwise the left-hand side of the original equation is even and the right-hand side is odd - say $k=2r$. Then divide through by $2$ and note that $r$ is likewise even, say $r=2l$.