It seems that you forgot something in the statement. What is the hypothesis about $n_1^2 + ... n_{1998}^2$ ? Pierre.

I believe it is that $n_1^2+...+n_{1997}^2=n_{1998}^2$ try looking at the equation mod 8.

Ooops! Yes,you're right,I've forgot smth.In fact n1 ² +...+n1997 ² =n1998 ² . But i dont understand what do you mean by studying the equation in mods 8,zcool.(i know what mods are) If you've got the solution,could you please post it ,with as much details as possible?

sorry about that, here is the solution in full 1) if at least two of the $n_1, n_2, ..., n_{1997}$ are even we would be done. 2) if one of the $n_1, n_2, ..., n_{1997}$ is even, then $n_1^2+n_2^2+...+n_{1997}^2$ must be even since we have an even number of odd terms in that sum, so $n_{1998}$ must be even and we have our two even numbers. for the sake of contradiction, assume that $n_1, n_2, ..., n_{1997}$ are odd. note that if x is odd, then $x^2 \equiv 1 \mod 8$, so $n_1^2+...+n_{1997}^2 \equiv 1997 \equiv 5 \equiv n_{1998}^2 \mod 8$ this is a contradiction because 5 is not a square mod 8. Therefore at least one of $n_1, n_2, ..., n_{1997}$ is even and from 1) and 2) we see that there are at least two even numbers.

Zscool's solution does not make any sense for the case in which one of them is even.

Actually, it makes perfect sense. If ony one is even, then both sides must be even, as an even of odd numbers always adds up to an even number. Then only one term on the LHS is even and the term on the RHS is even, meaning that two total are even.

If all numbers are uneve then n1^2+n2^2+...+n1997^2=M8+5=n1998^2=M8+1(F) If there is a single even number then :if(n1988-even then in n1^2+n2^2+...+n1997^2 there is a number even) if one a number from n1,n2,n3,..,n1997 then n1998 is even then two number even. So there are at least 2 number even.

Suppose that all the numbers on the LHS are odd. If a odd number is a perfect square it has to be 1 (mod 8) because $1^2\equiv3^2\equiv5^2\equiv7^2\equiv1$$(\text{mod}$ $8)$. Thus the LHS is congruent to 5 (mod 8). But 5 is a quadratic non-residue modulo 8, which means we have a contradiction. Now suppose that we have one number on the LHS that is even. Any odd perfect square modulo 4 is congruent to 1 (mod 4) because $1^2\equiv3^2\equiv1$$(\text{mod}$ $4)$. Thus the LHS is congruent to 0 (mod 4) which means that the RHS is also congruent to 0 (mod 4) implying that $n_{1998}$ is even. Thus we are done.

It is trivial that the condition is true when RHS is even. And if the RHS is odd, then LHS have to contain even number of even number. So, only hard case is to proof that it is impossible to have all integer odd in LHS. We know that every odd number can be written as $2k+1$ form. So every square of a odd number can be written as $(2k+1)^2=4k^2+4k+1$. Now, we can write LHS as $$ \displaystyle\sum_{i=1}^{1997} 4k_i^2 +\sum_{i=1}^{1997}4k_i+1997=4k^2_{1998}+4k_{1998}+1 $$ $$ \Rightarrow \displaystyle\sum_{i=1}^{1997} 4k_i^2 +\sum_{i=1}^{1997}4k_i+1996=4k^2_{1998}+4k_{1998} $$ Dividing by $4$ we get, $$ \displaystyle\sum_{i=1}^{1997} k_i^2 +\sum_{i=1}^{1997}k_i+499=k^2_{1998}+k_{1998} $$ Now,$\displaystyle\sum_{i=1}^{1997} k_i^2$ and $\displaystyle\sum_{i=1}^{1997} k_i$ has same parity and $499$ is odd. So LHS is odd. But $4k^2_{1998}+4k_{1998}$ is even. $Contradiction$. So there has to be more even numbers of even numbers in LHS. $[Done]$