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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Number theory Difficulty 5.1 AIME, harder Prove it Romania

Show that there is an infinite number of positive integers tt such that none of the equations x2+y6=tx^2 + y^6 = t, x2+y6=t+1x^2 + y^6 = t + 1, x2−y6=tx^2 - y^6 = t, x2−y6=t+1x^2 - y^6 = t + 1 has solutions (x,y)∈Z×Z(x, y) \in \mathbb{Z} \times \mathbb{Z}.

Solution

If xx is a positive integer, then either x12≡0(mod13)x^{12} \equiv 0 \pmod{13} or x12≡1(mod13)x^{12} \equiv 1 \pmod{13}, hence x6x^6 is congruent with −1-1, 00 or 11 modulo 1313. Therefore, if tt is congruent with 6(mod13)6 \pmod{13}, then ±x6+t\pm x^6 + t is congruent with 55, 66 or 7(mod13)7 \pmod{13}, while ±x6+t+1\pm x^6 + t + 1 is congruent with 66, 77 or 8(mod13)8 \pmod{13}.

On the other hand, perfect squares are congruent with 00, 11, 44, 99, 33, 1212 or 10(mod13)10 \pmod{13}. Therefore a perfect square can not be equal to a number of the form ±x6+t\pm x^6 + t or ±x6+t+1\pm x^6 + t + 1 if t≡6(mod13)t \equiv 6 \pmod{13}. In conclusion, all the numbers that are congruent with 66 modulo 1313 have the required property.

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