Solve ## 4x+51y=9: x=15+51t, y=-1-4t ##

[Math100]

Homework Statement

Using congruences, solve the Diophantine equation below:
$4x+51y=9$.
[Hint: $4x\equiv 9\pmod {51}$ gives $x=15+51t$, whereas $51y\equiv 9\pmod {4}$ gives $y=3+4s$. Find the relation between $s$ and $t$.]

Relevant Equations

None.

Consider the Diophantine equation $4x+51y=9$.
Observe that $gcd(51, 4)=1$.
Then $4x\equiv 9\pmod {51}\implies 52x\equiv 117\pmod {51}\implies x\equiv 15\pmod {51}$.
Now we have $51y\equiv 9\pmod {4}\implies 3y\equiv 1\pmod {4}\implies y\equiv 3\pmod {4}$.
This means $x=15+51t$ and $y=3+4s, \forall t, s$.
Since $4(15+51t)+51(3+4s)=9$, it follows that $s=-1-t$.
Thus $y=3+4s=3+4(-1-t)=-1-4t$.
Therefore, $x=15+51t$ and $y=-1-4t, \forall t, s$.

 

[fresh_42]

Homework Statement:: Using congruences, solve the Diophantine equation below:
$4x+51y=9$.
[Hint: $4x\equiv 9\pmod {51}$ gives $x=15+51t$, whereas $51y\equiv 9\pmod {4}$ gives $y=3+4s$. Find the relation between $s$ and $t$.]
Relevant Equations:: None.

Consider the Diophantine equation $4x+51y=9$.
Observe that $gcd(51, 4)=1$.
Then $4x\equiv 9\pmod {51}\implies 52x\equiv 117\pmod {51}\implies x\equiv 15\pmod {51}$.
Now we have $51y\equiv 9\pmod {4}\implies 3y\equiv 1\pmod {4}\implies y\equiv 3\pmod {4}$.
This means $x=15+51t$ and $y=3+4s, \forall t, s$.
Since $4(15+51t)+51(3+4s)=9$, it follows that $s=-1-t$.
Thus $y=3+4s=3+4(-1-t)=-1-4t$.
Therefore, $x=15+51t$ and $y=-1-4t, \forall t, s$.

So far so good, except that you should toss the "s" in the last line since we only have one parameter $t$ left.

Theoretically, you must check whether your solution is actually one. You derived necessary conditions for your solution and got a set of possible solutions. Now, we check whether they are sufficient, too.
\begin{align*}
4x+51y= 9 &\Longleftrightarrow 4\cdot (15+51t)+51\cdot (-1-4t)=60-51=9
\end{align*}

This has - strictly speaking - always to be done if the steps of a calculation cannot be reversed, e.g. taking roots or squaring numbers, or if we use congruences. So either we could write $\Longleftrightarrow$ along every step of a proof, or we deduce solutions and check whether they fulfill our requirement. The latter is usually easier to do.

 

[WWGD]

Or notice 51=13(4)-1. From there, move terms ahead and multiply, for the first half. Then you can generalize to all solutions other than 4 , and choose one that is 15(Mod 51).