I solved the exercise 2 of this pdf of Halmos:
http://math.slu.edu/~srivastava/Halmos.pdf
Show that $a^{2} + b^{2} = 10a+b$ has no solutions if $a$ and $b$ are digits and $a \neq 0$.
Suppose that the equation has a solution. Note that $a \in \{1,2,3,4,5,6,7,8,9\}$ and $b \in \{0,1,2,3,4,5,6,7,8,9\}$
$a²+b² = 10a+b$
$a²-10a = -b²+b = -b(b-1)$
$a(a-10) = -b(b-1)$
$-a(a-10) = b(b-1)$
On the LHS, $-a(a-10)$ is a strictly negative number, since $a\neq 0$ is positive and $(a-10)$ is negative for any possible value of $a$.
If $b=0$ or $b=1$, the RHS is zero. Absurd.
If $b \neq 0$ and $b \neq 1$, the RHS is a strictly positive number, since $b$ is positive and $(b-1)$ is positive. Absurd.
Then there is no solution. $$ \square $$
I want to know if my proof is alright and how other ways can I prove this statement.
Thanks.
As Brian said, $−a(a−10)$ and $b(b−1)$ are both strictly positive so there is no contradiction. A way I went about was similar although less elegant. I noticed that in the above expression, it is possible for the RHS to be greater or lesser than the LHS so any proof based on a contradiction of the sign of both sides will not work. So,
$a(10-a)=b(b-1)$
Notice that the RHS is always even. So for the LHS to be even, $a$ must be even. Further, the maximum value of $a(10-a)$ occurs when $a=5$ and is equal to $25$. Thus $b \le 5$, so $b(b-1) < 25 $. Now, we have cut down the possible cases for $a$ and $b$.
$a\in\{2,4,6,8\}$ and $b\in\{1,2,3,4,5\}$
Let $a=2m$ with $m={1,2,3,4}$ and $b(b-1) = 2k$ with $k=0,1,3,6,10$. Therefore
$2m(10-2m)=2k$
$4m(5-m)=2k$
$2m(5-m)=k$
Now the LHS is always a multiple of a $4$. However, $k$ is not a perfect multiple of $4$ for any value.
Thus, we reach a contradiction. Hence, the equation cannot have a root satisfying the given conditions.
Whew...Please point out any mistakes ! Thanks.