Friend squares numbers 
Two perfect squares are friends if one is obtained from the other
  adding the digit $1$ at the left.
For instance, $1225 = 35^2$ and $225 = 15^2$ are friends.
Prove that there are infinite pairs of odd perfect squares that are
  friends.

Solution: Suppose that $n^2$ and $m^2$ are friends and both odd, with the former being greater than the latter. Then $n$ and $m$ are both odd, and $n^2 - m^2$ is a power of ten.
Then, because of the difference-of-squares identity $n^2 - m^2 = (n-m)(n+m)$, this means that $n+m = 2^a 5^b$ for some $a, b$ and $n-m = 2^c 5^d$ for some $c, d$, where $a + c = b + d$.
Adding the two equations, we obtain $2n = 2^a 5^b + 2^c 5^d$. The right hand side has to be even, but not a multiple of 4. This means that $a$ and $c$ have to be both positive or zero. If they are both positive, they cannot be both at least two (since the right hand side would be a multiple of 4), but they cannot be both one (since $2^a 5^b + 2^c 5^d = 2(5^b + 5^d)$ would be a multiple of 4), so one of them has to be one and the other has to be at least two. If $a$ and $c$ are both zero, then $2^a 5^b + 2^c 5^d = 5^b + 5^d \equiv 1^b + 1^d \equiv 2\pmod{4}$, so all such cases work.
Similarly, subtracting the two equations, we obtain $2m = 2^a 5^b - 2^c 5^d$. The right hand side has to be even, but not a multiple of 4. This means that $a$ and $c$ have to be both positive or both zero. If they are both positive, they cannot be both at least two, and they cannot be both one, so one of them has to be one and the other has to be at least two. However, if $a$ and $c$ are both zero, then $2^a 5^b - 2^c 5^d = 5^b - 5^d \equiv 1^b - 1^d \equiv 0\pmod{4}$, so no such cases work.
In conclusion, we have demonstrated that $n^2$ and $m^2$ are friends and both odd iff $n+m = 2^a 5^b$ for some $a, b$ and $n-m = 2^c 5^d$ for some $c, d$, where $a + c = b + d$ and where $a = 1, c \geq 2$ or vice versa. This clearly covers infinitely many cases, so we are done.

Why is $ n ^ 2 - m ^ 2 $ a power of ten?


You may be interested in a relatively simple method of generating an infinite sequence of such numbers.

First we shall prove that if $A$ and $B$ are powers of $5$ and $C$ a power of $2$ such that
$\,\,\,\,\,\,2A^2BC$ is a power of $10$ and $1\le\frac{C}{B}\le \frac{5}{2},$
then a solution is given by $n=A(5B+C)$ and $m=A(5B-C)$.

$n^2-m^2=20A^2BC$ is a power of $10$ and so we require $$20A^2BC>A^2(5B-C)^2\ge 2A^2BC.$$ This gives two inequalities which are both satisfied when $\frac{C}{B}$ is in the above range $$C^2-12BC+25B^2\ge 0 \text {   and  } C^2-30BC+25B^2<0. $$

$A=5,B=1,C=2$ is an example of numbers satisfying the condition given above. From any such solution we can generate a larger solution as follows.

If  $\frac{C}{B}< \frac{25}{16}$ then $(A',B',C')=(5A,5B,8C)$
If  $\frac{C}{B}\ge \frac{25}{16}$ then $(A',B',C')=(5A,25B,16C)$

The first few solutions generated from $(5,1,2)$ are as follows.
$n^2=1225$ and $m^2=225$
$n^2=15405625$ and $m^2=5405625$
$n^2=12127515625$ and $m^2=2127515625.$

