matqkks said:
Are there any real life applications of linear Diophantine equations? I am looking for examples which will motivate students.
I do hope that Your students will be 'motivated' by the following 'brillant' application od diophantine equations that is datec from the Middle Ages. A well known fundamental theorem of the number theory is called 'Chinese Remainder Theorem' and it extablishes that if $n_{1}$ and $n_{2}$ are coprime, then the diophantine equation...
$\displaystyle x \equiv a_{1}\ \text{mod}\ n_{1}$ $\displaystyle x \equiv a_{2}\ \text{mod}\ n_{2}\ (1)$... has one and only one solution $\text{mod}\ n_{1}\ n_{2}$. It is easy to demonstrate in a more general case that if $n_{1},\ n_{2},\ ...\ n_{k}$ are coprime, then the diophantine equation... $\displaystyle x \equiv a_{1}\ \text{mod}\ n_{1}$
$\displaystyle x \equiv a_{2}\ \text{mod}\ n_{2}$
$\displaystyle ...$
$\displaystyle x \equiv a_{k}\ \text{mod}\ n_{k}\ (2)$
... has one and one solution $\text{mod}\ N= n_{1}\ n_{2}\ ...\ n_{k}$. All that is well known but may be it is not as well known why this theorem is called 'chinese'. The reason seems to be in the fact that in the old China the mathematical knowledge was 'patrimony' of the highest social classes and the rest of population was able to count till twenty and no more. Taking into account that, when a chinese general wanted to know the number of soldiers of one batalion he instructed the commander to marshal the soldiers first in rows of 7, then in rows of 11 and then in rows of 13 and any time to count the soldiers in the last row. The unknown number of soldiers can be ontained solving the diphantine equation (2) where $n_{1}=7,\ n_{2}= 11,\ n_{3}=13$ so that $N=n_{1}\ n_{2}\ n_{3}=1001$. The general procedure to solve (2) is the following... a) we define for i=1,2,...,k $\displaystyle N_{i}= \frac{N}{n_{i}}$ and $\displaystyle \lambda_{i} \equiv N_{i}^{-1}\ \text{mod}\ n_{i}$ b) we compute directly...
$\displaystyle x = a_{1}\ \lambda_{1}\ N_{1} + a_{2}\ \lambda_{2}\ N_{2} + ...+ a_{k}\ \lambda_{k}\ N_{k}\ \text{mod}\ N\ (3)$
In the case of chinese generals is $\displaystyle N_{1}= 143,\ \lambda_{1} \equiv 5\ \text{mod} 7,\ N_{2}= 91,\ \lambda_{2} \equiv 4\ \text{mod}\ 11,\ N_{3}= 77,\ \lambda_{3} \equiv 12\ \text{mod}\ 13$ so that is... $\displaystyle x \equiv 715\ a_{1} + 364\ a_{2} + 924\ a_{3}\ \text{mod}\ 1001\ (4)$
Kind regards
$\chi$ $\sigma$