อ้างอิง:
ข้อความเดิมเขียนโดยคุณ หยินหยาง
hint : $a^4+4b^4 =(a^2-2ab+2b^2)(a^2+2ab+2b^2)$
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เมื่อเช้า ผมคิดแนวนี้ แต่ก็ไปต่อไม่ถูก
ก็เลยหันมาใช้แบบตัวเลขแทน ถึกดี
จาก $x^2+y^2 = (x+y)^2 - 2xy$
จัดรูปข้างต้นเป็น
$(7^2)^2 +(2^3)^2 = (49)^2+ (8)^2 = (49+8)^2 - 2\cdot 8 \cdot 49 = 57^2-28^2 = (57+28)(57-28) = (85)(29)$
$(15^2)^2 +(2^3)^2 = (225)^2+ (8)^2 = (225+8)^2 - 2\cdot 8 \cdot 225 = 233^2-60^2 = (233+60)(233-60) = (293)(173)$
$(23^2)^2 +(2^3)^2 = (529)^2+ (8)^2 = (529+8)^2 - 2\cdot 8 \cdot 23^2 = 537^2-92^2 = (537+92)(537-92) = (629)(445)$
$(31^2)^2 +(2^3)^2 = (961)^2+ (8)^2 = (961+8)^2 - 2\cdot 8 \cdot 31^2 = 969^2-124^2 = (969+124)(969-124) = (1093)(845)$
$(39^2)^2 +(2^3)^2 = (1521)^2+ (8)^2 = (1521+8)^2 - 2\cdot 8 \cdot 39^2 = 1529^2-156^2 = (1529+156)(1529-156) = (1685)(1373)$
$(47^2)^2 +(2^3)^2 = (2209)^2+ (8)^2 = (2209+8)^2 - 2\cdot 8 \cdot 47^2 = 2217^2-188^2 = (2217+188)(2217-188) = (2405)(2029)$
$(3^2)^2 +(2^3)^2 = (9)^2+ (8)^2 = (9+8)^2 - 2\cdot 8 \cdot 9 = 17^2-12^2 = (17+12)(17-12) = (29)(5)$
$(11^2)^2 +(2^3)^2 = (121)^2+ (8)^2 = (121+8)^2 - 2\cdot 8 \cdot 11^2 = 129^2-44^2 = (129+44)(129-44) = (173)(85)$
$(19^2)^2 +(2^3)^2 = (361)^2+ (8)^2 = (361+8)^2 - 2\cdot 8 \cdot 19^2 = 369^2-76^2 = (369+76)(369-76) = (445)(293)$
$(27^2)^2 +(2^3)^2 = (729)^2+ (8)^2 = (729+8)^2 - 2\cdot 8 \cdot 27^2 = 737^2-108^2 = (737+108)(737-108) = (845)(629)$
$(35^2)^2 +(2^3)^2 = (1225)^2+ (8)^2 = (1225+8)^2 - 2\cdot 8 \cdot 35^2 = 1233^2-140^2 = (1233+140)(1233-140) = (1373)(1093)$
$(43^2)^2 +(2^3)^2 = (1849)^2+ (8)^2 = (1849+8)^2 - 2\cdot 8 \cdot 43^2 = 1857^2-172^2 = (1857+172)(1857-172) = (2029)(1685)$
$\dfrac{(85)(29)(293)(173)(629)(445)(1093)(845)(1685)(1373) (2405)(2029)}{ (29)(5)(173)(85)(445)(293)(845)(629)(1373)(1093)(2029)(1685)} = 481 $
มีใครถึกกว่านี้ไหม