Use the Euclidean Algorithm to find $a, b, c, d$ such that $225a + 360b +432c +480d = 3$ I wish to find the integers of $a,b,c$ and $d$ such that: 
$$225a + 360b +432c +480d = 3$$
which is equal to: 
$$75a + 120b +144c+ 160d =1$$
I know I have to use the Euclidean algorithm. And I managed to do it for two integers $x$ and $y$. But can't figure out, how to do it with $4$ integers.
 A: $$ 144 \cdot 12 - 75 \cdot 23 = 3 $$
$$ 160 \cdot 1 - 3 \cdot 53 = 1 $$
$$  160 \cdot 1 - 53 ( 144 \cdot 12 - 75 \cdot 23 ) =1 $$
$$  160 \cdot 1 - 636 \cdot 144 + 1219 \cdot 75 = 1  $$
The shortest vector solution is
$$   a= 3, b= -2, c= -1, d= 1  $$
with 
$$ 3 \cdot 75 - 2 \cdot 120 - 144 + 160 = 225 -240 -144 + 160 = 1   $$
The more interesting question is finding a basis for the lattice of integer vectors orthogonal to your vector. A basis is given by these three rows:
$$
\left(
\begin{array}{rrrr}
-175536 & 0 & 91585 & -144 \\
-146280 & 1 & 76320 & -120 \\
 -91424 & 0 & 47700 &  -75 \\
\end{array}
\right)
$$
This three dimensional lattice has Gram determinant $66361$
Next is a reduced basis by the LLL algorithm.
$$
\left(
\begin{array}{rrrr}
0 & 4 & 0 & -3 \\
0 & 2 & -5 & 3 \\
 8 & -1 & 0 &  -3 \\
\end{array}
\right)
$$
The Gram matrix for the reduced basis, still with determinant 66361, is
$$
\left(
\begin{array}{rrr}
25 & -1 & 5  \\
-1 & 38 & -11  \\
 5 & -11 & 74  \\
\end{array}
\right)
$$
There is a theorem involved here,
$$ 75^2 + 120^2 + 144^2 + 160^2 = 66361 $$
All solutions $(a,b,c,d)$ are given, with integer $s,t,u,$ by
$$ (3+8u,-2+4s+2t-u,-1-5t,1-3s+3t-3u) $$
A: First divide both sides by $3$ to get $$ 75a+120b+144c+160d=1$$
Now let $$a=b=c=x$$
Your equation changes to $$339x+160d=1$$
You can solve this one for $x$ and $d$ because $339$ and $160$ are relatively prime. 
Back substitute and you have your solution. 
A: $\color{#c00}6 = 2(75)\!-\!144,\, $ $\color{#0a0}{80} = 2(120)\!-\!160\ $ so $\,\bbox[5px,border:1px solid red]{1 = 75\!+\!\color{#c00}6\!-\!\color{#0a0}{80} = 3(75)-2(120) -144 + 160}$
Remark $ $ Found by perusing coef remainders: $\bmod 75:\ 144\equiv -\color{#c00}6,\,$ $\,\bmod 120\!:\ 160\equiv -\color{#0a0}{80}$ 
i.e. we applied a few (judicious) steps of the extended Euclidean algorithm. 
Alternatively, applying the algorithm mechanically, reducing each argument by the least argument, we get a longer computation like that in user21820's answer, viz.
$$\begin{align} 
&\gcd(\color{#88f}{75},120,144,160)\ \  {\rm so\ reducing}\ \bmod \color{#8af}{75}\\
=\ &\gcd (75,\ 45,\,-\color{#0a0}6,\ \ 10)\ \ \ {\rm so\ reducing}\ \bmod \color{#0a0}{6} \\
=\ &\gcd(\ 3,\ \ \ \ 0,\ {-}\color{#0a0}6,\ {-}\color{#f4f}2)\ \ \ {\rm so\ reducing}\ \bmod \color{#F4f}{2}\\
=\ &\gcd(\ \color{#d00}{\bf 1},\ \ \ \ 0,\ \ \ \ 0,\ {-}2)\\
\end{align}\qquad\qquad $$
yielding a Bezout identity for $\color{#d00}{\bf 1}$ from the augmented matrix in the extended algorithm (follow the above link for a complete presentation with the augmented matrix displayed).
A: You can systematically solve any such equation (or prove that there are no solutions) by the following:
Take any integers $a,b,c,d$. Then the following correspond:


*

*Solutions of $75a+120b+144c+160d = 1$

*Solutions of $120b+144c+160d \equiv 1 \pmod{75}$

*Solutions of $45b-6c+10d \equiv 1 \pmod{75}$

*Solutions of $45b-6c+10d+75p = 1$ where $p$ is an integer

*Solutions of $45b+10d+75p \equiv 1 \pmod{6}$

*Solutions of $3b+4d+3p \equiv 1 \pmod{6}$

*Solutions of $3b+4d+3p+6q = 1$ where $q$ is an integer

*Solutions of $4d \equiv 1 \pmod{3}$
And now you simply follow the reverse correspondences.
