# one equation three unknowns

Mathematics Asked by Stacey on October 21, 2020

Could someone point me in the right direction as to how to solve:

7a+8b+9c = 1744


I need to know how to do this for other three unknown equations a well so the answer alone won’t help me.

Thanks

The set of integer $$a,b,c$$ such that $$7a+8b+9c=0$$ is called a lattice, on this site the tag is integer-lattices. https://math.stackexchange.com/questions/tagged/integer-lattices

Next, we find a basis for the lattice. We do this by applying column operations as multiplication on the right by elementary matrices.

$$left( begin{array}{rrr} 7&8&9 \ end{array} right) left( begin{array}{rrr} 1&0&0 \ 7&9&-1 \ -7&-8&1 \ end{array} right) = left( begin{array}{rrr} 0&0&1 \ end{array} right)$$

The outcom of doing this is that the $$0$$ positions in the final row $$(0,0,1)$$ mark the basis vectors, as columns. So, a basis is

$$left( begin{array}{rr} 1&0 \ 7&9 \ -7&-8 \ end{array} right)$$ At this stage it is more convenient to write as rows: the entire lattice is parametrized as $$x(1,7,-7) + y ( 0,9,-8)$$ or $$( x, 7x+9y, -7x-8y)$$

All that remains is to find a single solution for $$1744$$ and add these. Factoring reveals $$1744 = 16 cdot 109 = 8 cdot 218,$$ so a fixed solution is $$( 0, 218,0)$$

and the set of all solutions is parametrized over the integers as

$$color{blue}{( x, 218+ 7x+9y, -7x-8y)}$$

If I had to obey inequalities I would define $$t = -y$$ and write

$$color{magenta}{( x, 218+ 7x-9t, -7x+8t)}$$ Using these (integer) variables, the non-negative solutions of the original problem occur in the narrow triangle $$x geq 0, ; ; ; 9t-218 leq 7x leq 8t$$

In this dimension one may calculate a reduced basis by ordinary Gauss reduction of the binary quadratic form. The result is a different parametrization, I should use different letters

$$color{red}{( -u+4v, 218+ 2u+v, -u-4v)}$$

If ( I cannot tell) you have inequalities (such as all entries non-negative) the first version is probably more convenient, as the $$x$$ is isolated.

$$bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc$$

Question about the matrix methodology: Column vector $$w$$ with element gcd = 1. The row is $$w^T.$$ We have any lattice element $$h ; , ;$$ meaning $$w^T h = 0.$$ And we construct a square integer unimodular matrix $$W$$ with $$w^T W = (1,0,0,0...,0).$$ This gives $$w^T =(1,0,0,...,0) W^{-1}.$$ Put together with $$w^T h = 0,$$ we arrive at $$(1,0,0,0...0) W^{-1} h = 0.$$ In turn, this means the first element in $$W^{-1}h$$ is zero, the others are a bunch of integers. so $$W^{-1} h = left( begin{array}{c} 0 \ g_2 \ g_3 \ vdots \ g_n \ end{array} right)$$ Multiply on the left by $$W$$ for $$h = W left( begin{array}{c} 0 \ g_2 \ g_3 \ vdots \ g_n \ end{array} right)$$ If you write this out, you see it means that $$h$$ itself is an integer linear combination of the columns of $$W$$ except the first (left hand) column. We get $$h = g_2 c_2 + g_3 c_3 + cdots + g_n c_n,$$ where $$c_j$$ is column $$j$$ in the square matrix $$W.$$ The second through final columns of $$W$$ really do span the lattice of integer vectors orthogonal to $$w^T.$$ Furthermore, $$det W = 1$$ says that they are independent; the second through $$n$$th columns of $$W$$ are a basis.

Answered by Will Jagy on October 21, 2020

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