1. Find all natural numbers $n$ that are less than 100 and have the property that the squares of the numbers $7n+5$ and $4n+3$ end in the same two-digit number.
1. Since the number $4 n+3$ is odd, the number $7 n+5$ must also be odd, so the number $n$ must be even: $n=2 k$ for a suitable integer $k$.

The required property can be expressed as follows: the difference $D=(7 n+5)^{2}-(4 n+3)^{2}$ is divisible by the number 100. Using the factorization

$$
D=((7 n+5)-(4 n+3))((7 n+5)+(4 n+3))=(3 n+2)(11 n+8)
$$

substituting $n=2 k$ gives the expression $D=4(3 k+1)(11 k+4)$. We are therefore interested in when the product $(3 k+1)(11 k+4)$ is divisible by 25. The factors $3 k+1$ and $11 k+4$ cannot both be multiples of five at the same time, because for their greatest common divisor we get

$$
\operatorname{gcd}(11 k+4,3 k+1)=\operatorname{gcd}(3 k+1,2 k+1)=\operatorname{gcd}(2 k+1, k)=\operatorname{gcd}(k, 1)=1
$$

We will therefore determine when $25 \mid 3 k+1$ and when $25 \mid 11 k+4$. From the expressions

$$
3 k+1=3(k-8)+25 \quad \text { and } \quad 11 k+4=11(k-11)+125
$$

we see that $25 \mid 3 k+1$ if and only if $k=25 t+8$, while $25 \mid 11 k+4$ if and only if $k=25 t+11$ (the letter $t$ denotes an integer in both cases). The sought numbers $n=2 k$ are therefore numbers of the forms $n=50 t+16$ and $n=50 t+22$, in the range from 1 to 99 these are precisely the numbers $16,22,66$ and 72.

Another solution. First, we determine the last digits of the numbers $(7 n+5)^{2}$ and $(4 n+3)^{2}$ depending on the last digit of the number $n$:

| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $7 n+5$ | 5 | 2 | 9 | 6 | 3 | 0 | 7 | 4 | 1 | 8 |
| $(7 n+5)^{2}$ | 5 | 4 | $\mathbf{1}$ | 6 | 9 | 0 | $\mathbf{9}$ | 6 | 1 | 4 |
| $4 n+3$ | 3 | 7 | 1 | 5 | 9 | 3 | 7 | 1 | 5 | 9 |
| $(4 n+3)^{2}$ | 9 | 9 | $\mathbf{1}$ | 5 | 1 | 9 | $\mathbf{9}$ | 1 | 5 | 1 |

(The calculation of the entire table can be halved if we realize beforehand, as in the previous solution, that $n$ must be even.) We see that the numbers $(7 n+5)^{2}$ and $(4 n+3)^{2}$ end with the same digit precisely when the number $n$ ends with the digit 2 or 6. Each sought $n<100$ is therefore either of the form $n=10 a+2$ or of the form $n=10 a+6$, where $a$ is an unknown digit. Even though it would suffice to test all $2 \cdot 10=20$ such numbers $n$, we choose a different approach.

(i) For $n=10 a+2$ we have

$$
\begin{aligned}
& (7 n+5)^{2}=(70 a+19)^{2}=4900 a^{2}+2660 a+361 \\
& (4 n+3)^{2}=(40 a+11)^{2}=1600 a^{2}+880 a+121
\end{aligned}
$$

We see that the number $(7 n+5)^{2}$ has the same digit in the tens place as the number $6 a+6$ has in the units place; the number $(4 n+3)^{2}$, on the other hand, has the same digit in the tens place as the number $8 a+2$ has in the units place. We are therefore looking for digits $a$ for which the difference $(8 a+2)-(6 a+6)=2(a-2)$ ends in
the digit zero; this is clearly true only for $a=2$ and $a=7$, corresponding to the solutions $n=22$ and $n=72$.

(ii) For $n=10 a+6$ we have

$$
\begin{aligned}
& (7 n+5)^{2}=(70 a+47)^{2}=4900 a^{2}+6580 a+2209 \\
& (4 n+3)^{2}=(40 a+27)^{2}=1600 a^{2}+2160 a+729
\end{aligned}
$$

This time, the number of tens in these numbers is the same as the number of units in the numbers $8 a$ and $6 a+2$. The difference $8 a-(6 a+2)=2(a-1)$ ends in the digit zero only for $a=1$ and $a=6$. The corresponding solutions are $n=16$ and $n=66$.

For a complete solution, award 6 points. For the observation that the number $n$ ends in the digit 2 or 6, award 2 points.