Elements of Algebra

406 ELEMENTS PART II. upon the analysis and deductions which this demonstration requires. If any one denied the proposition, it would be maintaining that there may be such values of $x$ and $y$, as will make $x^4 + y^4$ a square, in great numbers, notwithstanding there are none in small numbers. But it will be seen, that if $x$ and $y$ had satisfactory values, we should be able, however great those values might be, to deduce from them less values equally satisfactory, and from these, others still less, and so on. Since, therefore, we are acquainted with no value in small numbers, except the two cases already mentioned, which do not carry us any farther, we may conclude, with certainty, from the following demonstration, that there are no such values of $x$ and $y$ as we require, not even among the greatest numbers. The proposition shall afterwards be demonstrated, with respect to the *difference* of two biquadrates, $x^4 - y^4$, on the same principle. 205. The following consideration, however, must be attended to at present, in order to be convinced that $x^4 + y^4$ can only become a square in the self evident cases which have been mentioned. 1. Since we suppose $x$ and $y$ prime to each other, that is, having no common divisor, they must either both be odd, or one must be even, and the other odd. 2. But they cannot both be odd, because the sum of two odd squares can never be a square; for an odd square is always contained in the formula $4n + 1$; and, consequently, the sum of two odd squares will have the form $4n + 2$, which being divisible by $2$, but not by $4$, cannot be a square. Now, this must be understood also of two odd biquadrate numbers. 3. If, therefore, $x^4 + y^4$ must be a square, one of the terms must be even and the other odd; and we have already seen, that, in order to have the sum of two squares a square, the root of one must be expressible by $p^2 - q^2$, and that of the other by $2pq$; therefore, $x^2 = p^2 - q^2$, and $y^2 = 2pq$; and we should have $x^4 + y^4 = (p^2 + q^2)^2$. 4. Consequently, $y$ would be even, and $x$ odd; but since $x^2 = p^2 - q^2$, the numbers $p$ and $q$ must also be the one even, and the other odd. Now, the first, $p$, cannot be even; for if it were, $p^2 - q^2$ would be a number of the form $4n - 1$, or $4n + 3$, and could not become a square: therefore $p$ must be odd, and $q$ even, in which case it is evident, that these numbers will be prime to each other. 5. In order that $p^2 - q^2$ may become a square, or