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Let $n_1,n_2,\cdots ,n_{1998}$ be positive integers such that

$$\ n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2 $$

Show that at least two of the numbers are even.

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.
$[Done]$