Finally, we shall solve two more problems – the last problem is rather surprising since at first glance, it doesn’t appear to involve congruences.

Problem 4: Prove that if n is a perfect square, then .

*Solution* : this is rather simple: one could always write *n* = *m*^{2} and consider all possibilities of *m* modulo 4. But it’s easier to just consider *m* mod 2:

- If
*m*= 2*k*is even, then*m*^{2}= 4*k*^{2}is a multiple of 4. - If
*m*= 2*k*+1 is odd, then*m*^{2}= 4*k*^{2}+ 4*k*+ 1 is congruent to 1 mod 4. ♦

The next problem illustrates a common technique, called **infinite descent**.

Problem 5: Prove that the only solution of in integers is x = y = z = 0.

*Solution* : first it is clear that if *x* = 0 or *y* = 0, then we must have *x* = *y* = *z* = 0. Hence we assume that *x*, *y* and *z* are all non-zero. Taken modulo 5, we have

Hence, the only possibility for to be a multiple of 5 is . Thus *x* and *y* are both multiple of 5: write *x* = 5*x’* and *y* = 5*y’ *which gives , i. e. . But this in turn means *z* is a multiple of 5: *z* = 5*z’*, so we are back to the original equation:

Since *x*, *y* and *z* are non-zero, so are *x’*, *y’* and *z’*. But here the trouble begins: we can apply the same logic above and come to the conclusion that *x’*, *y’* and *z’* are also all multiples of 5, and so on, and so on. Thus, *x*, *y* and *z* are divisible by arbitrarily high powers of 5(!) which is clearly impossible if they are non-zero. ♦