Elements of Algebra

402 ELEMENTS PART II. If we take the upper sign, we find $3p^2 = 3$, and $p = 1$; whence $x = 5$: and if we adopt the other sign, we get a value of $p$, which being irrational, is of no use; it follows, therefore, that there is no square, except $25$, which has the property required. 194. Question 3. Required squares, which, multiplied by $5$, and added to $7$, may produce cubes; or it is required that $5x^2 + 7$ should be a cube. Let us first seek the cases in which $5x^2 + 7y^2$ becomes a cube. By Article 188, $a$ being equal to $5$, and $c$ equal $7$, we shall find that we must have $x = 5p^3 - 21pq^2$, and $y = 15p^2q - 7q^3$; so that in our example $y$ being $= \pm 1$, we have $15p^2q - 7q^3 = q(15p^2 - 7q^2) = \pm 1$; therefore $q$ must be a divisor of $1$; that is to say, $q = \pm 1$; consequently, we shall have $15p^2 - 7 = \pm 1$; from which, in both cases, we get irrational values for $p$: but from which we must not, however, conclude that the question is impossible, since $p$ and $q$ might be such fractions, that $y = 1$, and that $x$ would become an integer; and this is what really happens; for if $p = \frac{1}{2}$, and $q = \frac{1}{2}$, we find $y = 1$, and $x = 2$; but there are no other fractions which render the solution possible. 195. Question 4. Required squares in integer numbers, the double of which, diminished by $5$, may be a cube; or it is required that $2x^2 - 5$ may be a cube. If we begin by seeking the satisfactory cases for the formula $2x^2 - 5y^2$, we have, in the 188th Article, $a = 2$, and $c = -5$; whence $x = 2p^3 + 15pq^2$, and $y = 6p^2q + 5q^3$: so that, in this case, we must have $y = \pm 1$; consequently, $6p^2q + 5q^3 = q(6p^2 + 5q^2) = \pm 1$; and as this cannot be, either in integer numbers, or even in fractions, the case becomes very remarkable, because there is, notwithstanding, a satisfactory value of $x$; namely, $x = 4$; which gives $2x^2 - 5 = 27$, or equal to the cube of $3$. It will be of importance to investigate the cause of this peculiarity. 196. It is not only possible, as we see, for the formula $2x^2 - 5y^2$ to be a cube; but, what is more, the root of this cube has the form $2p^2 - 5q^2$, as we may perceive by making $x = 4$, $y = 1$, $p = 2$, and $q = 1$; so that we know a case in which $2x^2 - 5y^2 = (2p^2 - 5q^2)^3$, although the two factors of $2x^2 - 5y^2$, namely, $x\sqrt{2} + y\sqrt{5}$, and $x\sqrt{2} - y\sqrt{5}$, which, according to our method, ought to be the cubes of $p\sqrt{2} + q\sqrt{5}$, and of $p\sqrt{2} - q\sqrt{5}$, are not cubes; for, in our case, $x\sqrt{2} + y\sqrt{5} = 4\sqrt{2} + \sqrt{5}$; whereas