Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 3 - Subjective Problem 2

Find all integer solutions to 2x4+1=y22 x^4 + 1 = y^2.

This problem is a part of Tessellate S.T.E.M.S. (2019)

Note by Tessellate S.T.E.M.S. Mathematics
2 years, 6 months ago

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Comments

How do we get the solution

V G - 2 years, 6 months ago

Assume WLOG that x,yx,y are nonnegative. Note that (0,±1)(0, \pm 1) are solutions, and henceforth assume that x,yx,y are positive. We will show that there are no solutions with x,yx,y positive.

Write 2x4=y2−1=(y−1)(y+1).2x^4 = y^2-1 = (y-1)(y+1). Then yy must be odd and xx must be even, say x=2z,x=2z, so we can rewrite as 8z4=(y−12)(y+12). 8z^4 = \left( \frac{y-1}2 \right) \left( \frac{y+1}2 \right). Since y−12\frac{y-1}2 and y+12\frac{y+1}2 are consecutive integers, they are relatively prime. So they must both be equal to some perfect power of 22 times a fourth power. But one of them is odd. So this leads to two cases:

Case 1: y−1=2a4,y+1=16b4,y-1 = 2a^4, y+1 = 16b^4, where aa is odd. In this case, we get 2a4+2=16b4,2a^4+2 = 16b^4, and looking mod 88 shows that this equation has no solutions.

Case 2: y+1=2a4,y−1=16b4,y+1 = 2a^4, y-1 = 16b^4, where aa is odd. In this case, we get 2a4−2=16b4.2a^4-2 = 16b^4. Since aa is odd, write it as a=2k+1,a=2k+1, k≥0,k \ge 0, and get 2(a−1)(a+1)(a2+1)=16b4k(k+1)(2k2+2k+1)=b4 \begin{aligned} 2(a-1)(a+1)(a^2+1) &= 16b^4 \\ k(k+1)(2k^2+2k+1) &= b^4 \end{aligned} The integers k,k+1,2k2+2k+1k,k+1,2k^2+2k+1 are pairwise relatively prime, and their product is a fourth power, so they must all be fourth powers. But the only way that the nonnegative consecutive integers k,k+1k,k+1 can both be fourth powers is if k=0,k=0, which leads to the solution (0,1)(0,1) that we have already found. So there are no other solutions.

The conclusion is that the only integer solutions are (0,±1).(0,\pm 1).

Patrick Corn - 2 years, 5 months ago
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